Here we provide RD Sharma Class 12 Ex 22.9 Solutions Chapter 22 Differential Equations for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 22.9 Solutions Chapter 22 Differential Equations book pdf download. Now you will get step-by-step solutions to each question.

Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 22 |

Exercise | 22.9 |

Category | RD Sharma Solutions |

**RD Sharma Class 12 Ex 22.9 Solutions Chapter 22 Differential Equations**

### Solve the following differential equations:

### Question 1. x^{2}dy + y(x + y)dy = 0

**Solution:**

We have,

x^{2}dy + y(x + y)dy = 0

dy/dx = -y(x + y)/x^{2}

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = -vx(x + vx)/x^{2}

v + x(dv/dx) = -v – v^{2}

x(dv/dx) = -2v – v^{2}

On integrating both sides,

log|v/(v + 2)|^{1/2 }= -log|x/c|

v/(v + 2) = c^{2}/x^{2}

yx^{2 }= (y + 2x)c^{2} (Where ‘c’ is integration constant)

### Question 2. (dy/dx) = (y – x)/(y + x)

**Solution:**

We have,

(dy/dx) = (y – x)/(y + x)

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (vx – x)/(vx + x)

v + x(dv/dx) = (v – 1)/(v + 1)

x(dv/dx) = (v – 1)/(v + 1) – v

x(dv/dx) = (v – 1 – v^{2 }– v)/(v + 1)

x(dv/dx) = -(v^{2 }+ 1)/(v + 1)

On integrating both sides,

∫vdv/(v^{2}+1)+∫dv/(v^{2}+1)=-∫(dx/x)

(1/2)log|v^{2 }+ 1| + tan^{-1}(v) = log(c/x)

log|(y^{2 }+ x^{2})/x^{2}| + 2tan^{-1}(y/x) = log(c/x)^{2}

log(y^{2 }+ x^{2}) – log(x)^{2 }+ 2tan^{-1}(y/x) = log(c/x)^{2}

log(y^{2 }+ x^{2}) + 2tan^{-1}(y/x) = 2log(c) (Where ‘c’ is integration constant)

### Question 3. (dy/dx) = (y^{2 }– x^{2})/2yx

**Solution:**

We have,

(dy/dx) = (y^{2 }– x^{2})/2yx

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (v^{2}x^{2 }– x^{2})/2vx^{2}

v + x(dv/dx) = (v^{2 }– 1)/2v

x(dv/dx) = [(v^{2 }– 1)/2v] – v

x(dv/dx) = (v^{2 }– 1 – 2v^{2})/2v

x(dv/dx) = -(v^{2 }+ 1)/2v

On integrating both sides,

log|v^{2}+1| = -log(x) + log(c)

log|v^{2}+1| = log(c/x)

y^{2}/x^{2 }+ 1 = |c/x|

(x^{2 }+ y^{2}) = cx (Where ‘c’ is integration constant)

### Question 4. x(dy/dx) = (x + y)

**Solution:**

We have,

x(dy/dx) = (x+y)

(dy/dx) = (x+y)/x

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x + vx)/x

v + x(dv/dx) = (1 + v)

x(dv/dx) = 1

dv = (dx/x)

On integrating both sides,

∫dv = ∫(dx/x)

v = log(x) + c

y/x = log(x) + c

y = xlog(x) + cx (Where ‘c’ is integration constant)

### Question 5. (x^{2 }– y^{2})dx – 2xydy = 0

**Solution:**

We have,

(x^{2 }– y^{2})dx – 2xydy = 0

(dy/dx) = (x^{2 }– y^{2})/2xy

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x^{2 }– v^{2}x^{2})/2xvx

v + x(dv/dx) = (1 – v^{2})/2v

x(dv/dx) = [(1 – v^{2})/2v] – v

x(dv/dx) = (1 – 3v^{2})/2v

On integrating both sides,

-(1/3)log(1 – 3v^{2}) = log(x) – log(c)

log(1 – 3v^{2}) = -log(x)^{3 }+ log(c)

(x^{2 }– 3y^{2})/x^{2 }= (c/x^{3})

x(x^{2 }– 3y^{2}) = c (Where ‘c’ is integration constant)

### Question 6. (dy/dx) = (x + y)/(x – y)

**Solution:**

We have,

(dy/dx) = (x + y)/(x – y)

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x + vx)/(x – vx)

v + x(dv/dx) = (1 + v)/(1 – v)

x(dv/dx) = [(1 + v)/(1 – v)] – v

x(dv/dx) = (1 + v – v + v^{2})/(1 – v)

x(dv/dx) = (1 + v^{2})/(1 – v)

On integrating both sides,

∫dv/(v^{2 }+ 1) – ∫vdv/(v^{2 }+ 1) = ∫(dx/x)

tan^{-1}(v) – (1/2)log(v^{2 }+ 1) = log(x) + c

tan^{-1}(y/x) – (1/2)log(y^{2}/x^{2 }+ 1) = log(x) + c

tan^{-1}(y/x) – (1/2)log(y^{2 }+ x^{2}) + log(x) = log(x) + c

tan^{-1}(y/x) = (1/2)log(y^{2 }+ x^{2}) + c (Where ‘c’ is integration constant)

### Question 7. 2xy(dy/dx) = (x^{2 }+ y^{2})

**Solution:**

We have,

2xy(dy/dx) = (x^{2 }+ y^{2})

(dy/dx) = (x^{2 }+ y^{2})/2xy

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x^{2 }+ v^{2}x^{2})/2xvx

v + x(dv/dx) = (1 + v^{2})/2v

x(dv/dx) = [(1 + v^{2})/2v] – v

x(dv/dx) = (1 – v^{2})/2v

On integrating both sides,

-log(1 – v^{2}) = log(x) – log(c)

log(1 – v^{2}) = -log(x) + log(c)

1 – y^{2}/x^{2 }= (c/x)

(x^{2 }– y^{2}) = cx (Where ‘c’ is integration constant)

### Question 8. x^{2}(dy/dx) = x^{2 }– 2y^{2 }+ xy

**Solution: **

We have,

x^{2}(dy/dx) = x^{2 }– 2y^{2 }+ xy

(dy/dx) = (x^{2 }– 2y^{2 }+ xy)/x^{2}

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x^{2 }– 2v^{2}x^{2 }+ xvx)/2xvx

v + x(dv/dx) = (1 – 2v^{2 }+ v)/x^{2}

x(dv/dx) = (1 – 2v^{2 }+ v) – v

x(dv/dx) = (1 – 2v^{2})

dv/(1 – 2v^{2}) = (dx/x)

On integrating both sides,

dv/(1 – 2v^{2}) = ∫(dx/x)

(Where ‘c’ is integration constant)

### Question 9. xy(dy/dx) = x^{2 }– y^{2}

**Solution:**

We have,

xy(dy/dx) = x

^{2 }– y^{2}(dy/dx) = (x

^{2 }– y^{2})/xyIt is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x

^{2 }– v^{2}x^{2})/xvxv + x(dv/dx) = (1 – v

^{2})/vx(dv/dx) = [(1 – v

^{2})/v] – vx(dv/dx) = (1 – 2v

^{2})/vvdv/(1 – 2v

^{2}) = (dx/x)On integrating both sides,

∫vdv/(1 – 2v

^{2}) = ∫(dx/x)∫4vdv/(1 – 2v

^{2}) = 4∫(dx/x)-log(1 – 2v

^{2}) = 4log(x) – log(c)log(1 – 2v

^{2}) = log(c/x^{4})(1 – 2y

^{2}/x^{2}) = c/x^{4}(x

^{2}-2y^{2})/x^{2 }= c/x^{4}x

^{2}(x^{2 }– 2y^{2}) = c (Where ‘c’ is integration constant)

### Question 10. ye^{x/y}dx = (xe^{x/y }+ y)dy

**Solution:**

We have,

ye

^{x/y}dx = (xe^{x/y }+ y)dy(dy/dx) = (xe

^{x/y }+ y)/ye^{x/y}It is a homogeneous equation,

So, put x = vy (i)

On differentiating both sides w.r.t x,

dx/dy = v + y(dv/dy)

So,

v + y(dv/dy) = (vye

^{vy/y }+ y)/ye^{vy/y}v + y(dv/dy) = (ve

^{v }+ 1)/e^{v}y(dv/dy) = [(ve

^{v }+ 1)/e^{v}] – vy(dv/dy) = (ve

^{v }+ 1 – ve^{v})/e^{v}y(dv/dy) = (1/e

^{v})e

^{v}dv = (dy/y)On integrating both sides,

∫e

^{v}dv = ∫(dy/y)e

^{v }= log(y) + log(c)e

^{x/y }= log(y) + log(c) (Where ‘c’ is integration constant)

### Question 11. x^{2}(dy/dx) = x^{2 }+ xy + y^{2}

**Solution:**

We have,

x

^{2}(dy/dx) = x^{2 }+ xy + y^{2}dy/dx = (x

^{2 }+ xy + y^{2})/x^{2}It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x

^{2 }+ xvx + v^{2}x^{2})/x^{2}v + x(dv/dx) = (1 + v + v

^{2})x(dv/dx) = (1 + v + v

^{2}) – vdv/(1 + v

^{2}) = (dx/x)On integrating both sides,

∫dv/(1 + v

^{2}) = ∫(dx/x)tan

^{-1}(v) = log|x| + ctan

^{-1}(y/x) = log|x| + c (Where ‘c’ is integration constant)

### Question 12. (y^{2 }– 2xy)dx = (x^{2 }– 2xy)dy

**Solution:**

We have,

(y

^{2 }– 2xy)dx = (x^{2 }– 2xy)dy(dy/dx) = (y

^{2 }– 2xy)/(x^{2 }– 2xy)It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (v

^{2}x^{2 }– 2xvx)/(x^{2 }– 2xvx)v + x(dv/dx) = (v

^{2 }– 2v)/(1 – 2v)x(dv/dx) = [(v

^{2 }– 2v – v + 2v^{2})/(1 – 2v)]x(dv/dx) = 3(v

^{2 }– 1)/(1 – 2v)-(2v – 1)dv/(v

^{2 }– v) = 3(dx/x)On integrating both sides,

-∫(2v – 1)dv/(v

^{2 }– v) = 3∫(dx/x)-log|v

^{2 }– v| = 3log|x| – log|c|log|v

^{2 }– v| = log|c/x^{3}|(y

^{2}/x^{2 }– y/x) = (c/x^{3})(y

^{2 }– xy) = c/xx(y

^{2 }– xy) = c (Where ‘c’ is integration constant)

### Question 13. 2xydx + (x^{2 }+ 2y^{2})dy = 0

**Solution:**

We have,

2xydx + (x^{2 }+ 2y^{2})dy = 0

dy/dx = -(2xy)/(x^{2 }+ 2y^{2})

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = -(2xvx)/(x^{2 }+ 2v^{2}x^{2})

v + x(dv/dx) = -(2v)/(1 + 2v^{2})

x(dv/dx) = -[(2v)/(1 + 2v^{2})] – v

On integrating both sides,

Substituting (3v + 2v^{3}) = z

On differentiating both sides w.r.t x,

3(1 + 2v)dv = dz

(1 + 2v)dv = (dz/3)

(1/3)∫(dz/z) = -∫(dx/x)

(1/3)log|z| = -log|x| + log|c|

log|3v + 2v^{3}| = log|c/x|^{3}

3y/x + 2(y/x)^{3 }= (c/x)^{3}

(3yx^{2 }+ 2y^{3}) = c (Where ‘c’ is integration constant)

### Question 14. 3x^{2}dy = (3xy + y^{2})dx

**Solution:**

We have,

3x

^{2}dy = (3xy + y^{2})dx(dy/dx) = (3xy + y

^{2})/3x^{2}It is a homogeneous equation,

So put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (3xvx + v

^{2}x^{2})/3x^{2}v + x(dv/dx) = (3v + v

^{2})/3x(dv/dx) = [(3v + v

^{2})/3] – vx(dv/dx) = (3v + v

^{2 }– 3v)/33(dv/v

^{2}) = (dx/x)On integrating both sides,

3∫(dv/v

^{2}) = ∫(dx/x)-(3/v) = log|x| + c

-3x/y = log(x) + c (Where ‘c’ is integration constant)

### Question 15. (dy/dx) = x/(2y + x)

**Solution:**

We have,

(dy/dx) = x/(2y + x)

It is a homogeneous equation,

So, put y = vx (i)

*On differentiating both sides w.r.t x,*

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = x/(2vx + x)

v + x(dv/dx) = 1/(2v + 1)

x(dv/dx) = [1/(2v + 1)] – v

x(dv/dx) = (1 – 2v^{2 }– v)/(2v + 1)

(2v + 1)dv/(2v^{2 }+ v – 1) = -(dx/x)

On integrating both sides,

∫(2v + 1)dv/(2v^{2 }+ v – 1) = -∫(dx/x)

Solving by partial fraction,

A(v + 1) + B(2v – 1) = (2v + 1) (i)

Putting v = -1 and solve above equation,

A(0) + B(-3) = (-1)

B = (1/3)

Putting v = -(1/2) and solve equation (i),

A(3/2) + B(0) = 2

A = (4/3)

(3/2)log|2v – 1| + (1/3)log|v + 1| = -log|x| + log|c|

log|(2v – 1)^{2}(v + 1)| = -log|x|^{3 }+ log|c|

|(2v – 1)^{2}(v + 1)| = (c/x^{3})

(2y/x – 1)^{2}(y/x + 1) = (c/x^{3})

(2y – x)^{2}(x + y) = c (Where ‘c’ is integration constant)

### Question 16. (x + 2y)dx – (2x – y)dy = 0

**Solution:**

We have,

(x + 2y)dx – (2x – y)dy = 0

(dy/dx) = (x + 2y)/(2x – y)

It is a homogeneous equation,

So, put y = vx (i)

*On differentiating both sides w.r.t x,*

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x + 2vx)/(2x – vx)

v + x(dv/dx) = (1 + 2v)/(2 – v)

x(dv/dx) = [(1 + 2v)/(2 – v)] – v

x(dv/dx) = (1 + 2v – 2v + v^{2})/(2 – v)

(2 – v)dv/(1 + v^{2}) = (dx/x)

On integrating both sides,

∫(2 – v)dv/(1 + v^{2}) = ∫(dx/x)

2∫dv/(1 + v^{2}) – ∫vdv/(1 + v^{2}) = log|x| + log|c|

2tan^{-1}v – (1/2)∫2vdv/(1 + v^{2}) = log|x| + log|c|

2tan^{-1}v – log|1 + v^{2}|^{1/2 }= log|cx|

2tan^{-1}v = log|cx√(1 + v^{2})|

(Where ‘c’ is integration constant)

### Question 17.

**Solution:**

We have,

It is a homogeneous equation,

So, put y = vx (i)

*On differentiating both sides w.r.t x,*

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (vx/x) – √(v^{2}x^{2}/x^{2 }– 1)

v + x(dv/dx) = v – √(v^{2 }– 1)

x(dv/dx) = -√(v^{2 }– 1)

dv/√(v^{2 }– 1) = -(dx/x)

On integrating both sides,

∫dv/√(v^{2 }– 1) = -∫(dx/x)

log|v + √(v^{2 }– 1)| = -log|x| + log|c|

|v + √(v^{2 }– 1)| = (c/x)

y + √(y^{2 }– x^{2}) = c (Where ‘c’ is integration constant)

### Question 18. (dy/dx) = (y/x){log(y) – log(x) + 1}

**Solution:**

We have,

(dy/dx) = (y/x){log(y) – log(x) + 1}

(dy/dx) = (y/x){log(y/x) + 1}

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = v{log(v) + 1}

v + x(dv/dx) = vlog(v) + v

x(dv/dx) = vlog(v)

dv/vlogv = (dx/x)

On integrating both sides,

∫dv/vlogv = ∫(dx/x)

Let, logv = z

On differentiating both sides,dv/v = dz

∫(dz/z) = ∫(dx/x)

log|z| = log|x| + log|c|

z = xc

log|v| = xc

log|y/x| = xc (Where ‘c’ is integration constant)

### Question 19. (dy/dx) = (y/x) + sin(y/x)

**Solution:**

We have,

(dy/dx) = (y/x) + sin(y/x)

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = v + sin(v)

x(dv/dx) = sin(v)

dv/sin(v) = (dx/x)

On integrating both sides,

∫dv/sin(v) = ∫(dx/x)

∫cosec(v)dv = ∫(dx/x)

log|tan(v/2)| = log(x) + log(c)

log|tan(y/2x)| = log|xc|

tan(y/2x) = |xc| (Where ‘c’ is integration constant)

### Question 20. y^{2}dx + (x^{2 }– xy + y^{2})dy = 0

**Solution:**

We have,

y^{2}dx + (x^{2 }– xy + y^{2})dy = 0

(dy/dx) = -(y^{2})/(x^{2 }– xy + y^{2})

It is a homogeneous equation,

So, put y = vx (i)

*On differentiating both sides w.r.t x,*

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = -(v^{2}x^{2})/(x^{2 }– xvx + v^{2}x^{2})

v + x(dv/dx) = -(v^{2})/(1 – v + v^{2})

y(dv/dx) = [-(v^{2})/(1 – v + v^{2})] – v

On integrating both sides,

∫dv/(1 + v^{2}) – ∫dv/v = ∫(dx/x)

tan^{-1}(v) – log(v) = log(x) + log(c)

tan^{-1}(y/x) – log|y/x| = log(xc)

tan^{-1}(y/x) = log|(y/x)xc|

tan^{-1}(y/x) = log|yc|

(Where ‘c’ is integration constant)

### Question 21. [x√(x^{2 }+ y^{2}) – y^{2}]dx + xydy = 0

**Solution:**

We have,

[x√(x^{2 }+ y^{2}) – y^{2}]dx + xydy = 0

dy/dx = -[x√(x^{2 }+ y^{2}) – y^{2}]/xy

dy/dx = [y^{2 }– x√(x^{2 }+ y^{2})]/xy

It is a homogeneous equation,

So, put y = vx (i)

*On differentiating both sides w.r.t x,*

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = [v^{2}x^{2 }– x√(x^{2 }+ v^{2}x^{2})]/xvx

v + x(dv/dx) = [v^{2 }– √(1 + v^{2})]/v

x(dv/dx) = [v^{2 }– √(1 + v^{2})]/v – v

x(dv/dx) = -(√(1 + v^{2})/v

vdv/√(1 + v^{2}) = -(dx/x)

On integrating both sides,

∫vdv/√(1 + v^{2}) = -∫(dx/x)

(1/2)∫2vdv/√(1 + v^{2}) = -∫(dx/x)

Let, 1 + v^{2 }= z

*On differentiating both sides,*

2vdv = dz

(1/2)∫dz/√z = -∫(dx/x)

√z = -log|x| + log|c|

√(1 + v^{2}) = log|c/x|

√(x^{2 }+ y^{2})/x = log|c/x|

√(x^{2 }+ y^{2}) = xlog|c/x| (Where ‘c’ is integration constant)

### Question 22. x(dy/dx) = y – xcos^{2}(y/x)

**Solution:**

We have,

x(dy/dx) = y – xcos

^{2}(y/x)(dy/dx) = y/x – cos

^{2}(y/x)It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = v – cos

^{2}(v)x(dv/dx) = -cos

^{2}(v)dv/cos

^{2}(v) = -(dx/x)On integrating both sides,

∫dv/cos

^{2}(v) = -∫(dx/x)∫sec

^{2}vdv = -∫(dx/x)tan(v) = -log|x| + log|c|

tan(y/x) = log|c/x| (Where ‘c’ is integration constant)

### Question 23. (y/x)cos(y/x)dx – {(x/y)sin(y/x) + cos(y/x)}dy = 0

**Solution:**

We have,

(y/x)cos(y/x)dx – {(x/y)sin(y/x) + cos(y/x)}dy = 0

It is a homogeneous equation,

So, put y = vx (i)

*On differentiating both sides w.r.t x,*

dy/dx = v + x(dv/dx)

So,

x(dv/dx) = (v^{2}cosv – vsinv – v^{2}cosv)/(sinv + vcosv)

x(dv/dx) = -vsinv/(sinv + vcosv)

On integrating both sides,

∫[(sinv + vcosv)/vsinv]dv = -∫(dx/x)

∫(dv/v) + ∫(cotv)dv = -∫(dx/x)

log|v| + log|sinv| = -log|x| + log|c|

log|vsinv| = log|c/x|

(y/x)sin(y/x) = (c/x)

ysin(y/x) = c (Where ‘c’ is integration constant)

### Question 24. xylog(x/y)dx + {y^{2 }– x^{2}log(x/y)}dy = 0

**Solution:**

We have,

xylog(x/y)dx + {y^{2 }– x^{2}log(x/y)}dy = 0

It is a homogeneous equation,

So, put x = vy (i)

*On differentiating both sides w.r.t* y,

dx/dy = v + y(dv/dy)

So,

v + y(dv/dy) = (v^{2}logv – 1)/(vlogv)

y(dv/dy) = [(v^{2}logv – 1)/(vlogv)] – v

y(dv/dy) = (v^{2}logv – 1 – v^{2}logv)/vlogv

y(dv/dy) = -(1/vogv)

vlogvdv = -(dy/y)

On integrating both sides,

∫vlogvdv = -∫(dy/y)

logv∫vdv – ∫{d/dv(logv)∫vdv}dv}dv = -∫(dy/y)

(v^{2}/2)logv – (1/2)∫(1/v)(v^{2}/2)dv = -logy + logc

(v^{2}/2)logv – (1/2)∫vdv = -logy + logc

(v^{2}/2)logv – (v^{2}/4) + logy = log|c|

(v^{2}/2)[logv – 1/2] + logy = log|c|

v^{2}[logv – (1/2)] + logy^{ }= log|c|

(x^{2}/y^{2})[log(x/y) – (1/2)] + logy^{ }= log|c| (Where ‘c’ is integration constant)

### Question 25. (1 + e^{x/y})dx + e^{x/y}(1 – x/y)dy = 0

**Solution:**

We have,

(1 + e^{x/y})dx + e^{x/y}(1 – x/y)dy = 0

It is a homogeneous equation,

So, put x = vy (i)

*On differentiating both sides w.r.t *y,

dx/dy = v + y(dv/dy)

So,

y(dv/dy) = -[e^{v}(1 – v)/(1 + e^{v})] – v

y(dv/dy) = (-e^{v }+ ve^{v }– v – ve^{v})/(1 + e^{v})

y(dv/dy) = -(v + ve^{v})/(1 + e^{v})

[(1 + e^{v})/(v + e^{v})]dv = -(dy/y)

On integrating both sides,

∫[(1 + e^{v})/(v + e^{v})]dv = -∫(dy/y)

log|(v + e^{v})| = -log(y) + log(c)

log|(v + e^{v})| = log|c/y|

(x/y) + e^{x/y }= c/y

x + ye^{x/y }= c (Where ‘c’ is integration constant)

### Question 26. (x^{2 }+ y^{2})dy/dx = (8x^{2 }– 3xy + 2y^{2})

**Solution:**

We have,

(x^{2 }+ y^{2})dy/dx = (8x^{2 }– 3xy + 2y^{2})

(dy/dx) = (8x^{2 }– 3xy + 2y^{2})/(x^{2 }+ y^{2})

It is a homogeneous equation,

So, put y = vx (i)

*On differentiating both sides w.r.t* x,

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (8x^{2 }– 3xvx + 2v^{2}x^{2})/(x^{2 }+ v^{2}x^{2})

v + x(dv/dx) = (8 – 3v + 2v^{2})/(1 + v^{2})

x(dv/dx) = [(8 – 3v + 2v^{2})/(1 + v^{2})] – v

x(dv/dx) = (8 – 4v + 2v^{2 }– v^{3})/(1 + v^{2})

(1 + v^{2})dv/(8 – 4v + 2v^{2 }– v^{3}) = (dx/x)

On integrating both sides,

Using partial fraction,

(1 + v^{2}) = Av(2 – v) + B(2 – v) + C(4 + v^{2})

(1 + v^{2}) = 2Av – Av^{2 }+ 2B – Bv + 4C + Cv^{2}

(1 + v^{2}) = (C – A)v^{2 }+ (2A – B)v + (2B + 4C)

Comparing the co-efficient of both sides,

(C – A) = 1

(2A – B) = 0

(2B + 4C) = 1

Solving above equations,

A = -(3/8)

B = -(3/4)

C = (5/8)

(Where ‘c’ is integration constant)

### Question 27. (x^{2 }– 2xy)dy + (x^{2 }– 3xy + 2y^{2})dx = 0

**Solution:**

We have,

(x

^{2 }– 2xy)dy + (x^{2 }– 3xy + 2y^{2})dx = 0(dy/dx) = (x

^{2 }– 3xy + 2y^{2})/(2xy – y^{2})It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x

^{2 }– 3xvx + 2v^{2}x^{2})/(2xvx – x^{2})v + x(dv/dx) = (1 – 3v + 2v

^{2})/(2v – 1)x(dv/dx) = [(1 – 3v + 2v

^{2})/(2v – 1)] – vx(dv/dx) = (1 – 3v + 2v

^{2 }– 2v^{2 }+ v)/(2v – 1)x(dv/dx) = (1 – 2v)/(2v – 1)

x(dv/dx) = -1

dv = -(dx/x)

On integrating both sides,

∫dv = -∫(dx/x)

v = -log|x| + log|c|

(y/x) + log|x| = log|c| (Where ‘c’ is integration constant)

### Question 28. x(dy/dx) = y – xcos^{2}(y/x)

**Solution:**

We have,

x(dy/dx) = y – xcos

^{2}(y/x)(dy/dx) = y/x – cos

^{2}(y/x)It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = v – cos

^{2}(v)x(dv/dx) = -cos

^{2}(v)dv/cos

^{2}(v) = -(dx/x)On integrating both sides,

∫dv/cos

^{2}(v) = -∫(dx/x)∫sec

^{2}vdv = -∫(dx/x)tan(v) = -log|x| + log|c|

tan(y/x) = log|c/x| (Where ‘c’ is integration constant)

### Question 29. x(dy/dx) – y = 2√(y^{2 }– x^{2})

**Solution:**

We have,

x(dy/dx) – y = 2√(y^{2 }– x^{2})

(dy/dx) = [2√(y^{2 }– x^{2}) + y]/x

It is a homogeneous equation,

So, put y = vx (i)

*On differentiating both sides w.r.t x,*

dy/dx = v + x(dv/dx)

So,

x(dv/dx) = 2√(v^{2 }– 1)

dv/√(v^{2 }– 1) = 2(dx/x)

On integrating both sides,

∫dv/√(v^{2 }– 1) = 2∫(dx/x)

log|v + √(v^{2}– 1)| = 2log(x) + log(c)

|v + √(v^{2 }– 1)| = |cx^{2}|

(Where ‘c’ is integration constant)

### Question 30. xcos(y/x)(ydx + xdy) = ysin(y/x)(xdy – ydx)

**Solution:**

We have,

xcos(y/x)(ydx + xdy) = ysin(y/x)(xdy – ydx)

xycos(y/x)dx + x^{2}cos(y/x)dy = xysin(y/x)dy – y^{2}sin(y/x)dx

x^{2}cos(y/x)dy – xysin(y/x)dy = -y^{2}sin(y/x)dx – xycos(y/x)dx

It is a homogeneous equation,

So, put y = vx (i)

*On differentiating both sides w.r.t x,*

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (vcosv + v^{2}sinv)/(vsinv – cosv)

x(dv/dx) = [(vcosv + v^{2}sinv)/(vsinv – cosv)] – v

x(dv/dx) = (vcosv + v^{2}sinv – v^{2}sinv + vcosv)/(vsinv – cosv)

x(dv/dx) = (2vcosv)/(vsinv – cosv)

[(vsinv – cosv)/(vcosv)]dv = 2(dx/x)

On integrating both sides,

∫tanvdv – ∫(dv/v) = 2log|x| + log|c|

log|secv| – log|v| = log|cx^{2}|

log|(secv/v)| = log|cx^{2}|

(x/y)sec(y/x) = cx^{2}

sec(y/x) = cxy (Where ‘c’ is integration constant)

### Question 31. (x^{2 }+ 3xy + y^{2})dx – x^{2}dy = 0

**Solution:**

We have,

(x^{2 }+ 3xy + y^{2})dx – x^{2}dy = 0

dy/dx = (x^{2 }+ 3xy + y^{2})/x^{2}

It is a homogeneous equation,

So, put y = vx (i)

*On differentiating both sides w.r.t x,*

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x^{2 }+ 3xvx + v^{2}x^{2})/x^{2}

v + x(dv/dx) = (1 + 3v + v^{2})

x(dv/dx) = (1 + 3v + v^{2}) – v

x(dv/dx) = (1 + 2v + v^{2})

x(dv/dx) = (1 + v)^{2}

dv/(1 + v)^{2 }= (dx/x)

On integrating both sides,

∫dv/(1 + v)^{2 }= ∫(dx/x)

-[1/(v + 1)] = log|x| – c

x/(x + y) + log|x| = c (Where ‘c’ is an integration constant)

### Question 32. (x – y)(dy/dx) = (x + 2y)

**Solution:**

We have,

(x – y)(dy/dx) = (x + 2y)

(dy/dx) = (x + 2y)/(x – y)

It is a homogeneous equation,

So, put y = vx (i)

*On differentiating both sides w.r.t x,*

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x + 2vx)/(x – vx)

v + x(dv/dx) = (1 + 2v)/(1 – v)

x(dv/dx) = [(1 + 2v)/(1 – v)] – v

x(dv/dx) = (1 + 2v – v + v^{2})/(1 – v)

x(dv/dx) = (1 + v + v^{2})/(1 – v)

(1 – v)dv/(1 + v + v^{2}) = (dx/x)

On integrating both sides,

∫[(1 – v)/(1 + v + v^{2})]dv = ∫(dx/x)

(Where ‘c’ is an integration constant)

### Question 33. (2x^{2}y + y^{3})dx + (xy^{2 }– 3x^{2})dy = 0

**Solution:**

We have,

(2x^{2}y + y^{3})dx + (xy^{2 }– 3x^{2})dy = 0

dy/dx = (2x^{2}y + y^{3})/(3x^{3 }– xy^{2})

It is a homogeneous equation,

So, put y = vx (i)

*On differentiating both sides w.r.t x,*

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (2x^{2}vx + v^{3}x^{3})/(3x^{3 }– xv^{2}x^{2})

v + x(dv/dx) = (2v + v^{3})/(3 – v^{3})

x(dv/dx) = [(2v + v^{3})/(3 – v^{3})] – v

x(dv/dx) = (2v + v^{3 }– 3v + v^{3})/(3 – v^{3})

(3 – v^{3})dv/(2v^{3 }– v) = (dx/x)

On integrating both sides,

∫[(3 – v^{3})/(2v^{3 }– v)]dv = ∫(dx/x)

Using partial fraction,

3 – v^{2 }= A(2v^{2 }– 1) + (Bv + C)v

3 – v^{2 }= 2Av^{2 }– A + Bv^{2 }+ Cv

3 – v^{2 }= v^{2}(2A + B) + Cv – A

On comparing the coefficients, we get

A = -3,

B = 5,

C = 0,

-3log|v|+(5/4)log|2v^{2}-1|=log|x|+log|c|

-12log|v|+5log|2v^{2}-1|=4log|x|+4log|c|

|2y^{2 }– x^{2}|^{5 }= x^{2}c^{4}y^{12} (Where ‘c’ is an integration constant)

### Question 34. x(dy/dx) – y + xsin(y/x) = 0

**Solution:**

We have,

x(dy/dx) – y + xsin(y/x) = 0

x(dy/dx) = y – xsin(y/x)(dy/dx) = [y – xsin(y/x)]/x

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = [vx – xsinv]/x

v + x(dv/dx) = (v – sinv)

x(dv/dx) = -sinv

cosecvdv = -(dx/x)

On integrating both sides,

∫cosecvdv = -∫(dx/x)

-log|cosecv + cotv| = -log|x| + log|c|

-log|(1/sinv) + (cosv/sinv)| = -log|x/c|

|(1 + cosv)/sinv| = |x/c|

xsinv = c(1 + cosv)

xsin(y/x) = c[1 + cos(y/x)] (Where ‘c’ is integration constant)

### Question 35. ydx + {xlog(y/x)}dy – 2xydy = 0

**Solution:**

We have,

ydx + {xlog(y/x)}dy – 2xydy = 0

y + {xlog(y/x)}(dy/dx) – 2xy = 0

It is a homogeneous equation,

So, put y = vx (i)

*On differentiating both sides w.r.t x,*

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = v/(2 – logv)

x(dv/dx) = [v/(2 – logv)] – v

x(dv/dx) = (v – 2v + vlogv)/(2 – logv)

x(dv/dx) = -v(logv – 1)/(logv – 2)

On integrating both sides,

Let. logv – 1 = z

On differentiating both sides,

(dv/v) = dz

∫dz – ∫(dz/z) = -∫(dx/x)

z – log|z| = -log|x| + log|c|

(logv – 1) – log|(logv – 1)| = -log|x| + log|c|

logv – log|logv – 1| = -log|x| + log|c| + 1

log|(logv – 1)/v| = log|c_{1}x|

|logv – 1| = |c_{1}xv|

|log(y/x) – 1| = |c_{1}x(y/x)|

|log(y/x) – 1| = |c_{1}y| (Where ‘c_{1}’ is integration constant)

### Question 36(i). (x^{2 }+ y^{2})dx = 2xydy, y(1) = 0

**Solution:**

We have,

(x

^{2 }+ y^{2})dx = 2xydy(dy/dx) = (x

^{2 }+ y^{2})/2xyIt is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x

^{2 }+ v^{2}x^{2})/2vx^{2}v + x(dv/dx) = (1 + v

^{2})/2vx(dv/dx) = [(1 + v

^{2})/2v] – vx(dv/dx) = (1 + v

^{2 }– 2v^{2})/2vx(dv/dx) = (1 – v

^{2})/2v2vdv/(1 – v

^{2}) = (dx/x)On integrating both sides,

∫2vdv/(1 – v

^{2}) = ∫(dx/x)-log|1 – v

^{2}| = log|x| – log|c|log|1 – v

^{2}| = log|c/x||1 – y

^{2}/x^{2}| = |c/x||x

^{2 }– y^{2}| = |cx|At x = 1, y = 0

1 – 0 = c

c = 1

|x

^{2 }– y^{2}| = |x|(x

^{2 }– y^{2})^{ }= x

### Question 36(ii). xe^{x/y }– y + x(dy/dx) = 0, y(e) = 0

**Solution:**

We have,

xe

^{x/y }– y + x(dy/dx) = 0(dy/dx) = (y – xe

^{x/y})/xIt is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (vx – xe

^{v})/xv + x(dv/dx) = v – e

^{v}x(dv/dx) = v – e

^{v }– vx(dv/dx) = -e

^{v}e

^{-v}dv = -(dx/x)On integrating both sides,

∫e

^{-v}dv = -∫(dx/x)-e

^{-v }= -log|x| – log|c|e

^{-v }= log|x| + log|c|e

^{-(y/x) }= log|x| + log|c|At x = e, y = 0

e

^{-(0/e) }= log|e| + log|c|1 = 1 + log|c|

c = 0

e

^{-y/x }= logx

### Question 36(iii). (dy/dx) – (y/x) + cosec(y/x) = 0, y(1) = 0

**Solution:**

We have,

(dy/dx) – (y/x) + cosec(y/x) = 0

(dy/dx) = (y/x) – cosec(y/x)

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = v – cosec(v)

x(dv/dx) = v – cosec(v) – v

x(dv/dx) = -cosec(v)

-sin(v)dv = (dx/x)

On integrating both sides,

-∫sin(v)dv = ∫(dx/x)

cos(v) = log|x| + log|c|

cos(y/x) = log|x| + log|c|

At x = 1, y = 0

cos(0/1) = log|1| + log|c|

1 = 0 + log|c|

log|c| = 1

cos(y/x) = log|x| + 1

log|x| = cos(y/x) – 1

### Question 36(iv). (xy – y^{2})dx – x^{2}dy = 0, y(1) = 1

**Solution:**

We have,

(xy – y

^{2})dx – x^{2}dy = 0(dy/dx) = (xy – y

^{2})/x^{2}(dy/dx) = (y/x) – (y

^{2}/x^{2})It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = v – v

^{2}x(dv/dx) = v – v

^{2 }– vx(dv/dx) = -v

^{2}-(dv/v

^{2}) = (dx/x)On integrating both sides,

-∫(dv/v

^{2}) = ∫(dx/x)-(-1/v) = log|x| + c

(1/v) = log|x| + c

x/y = log|x| + c

At x = 1, y = 1

1 = log|1| + c

c = 1

x/y = log|x| + 1

y = x/[log|x| + 1]

### Question 36(v). (dy/dx) = [y(x + 2y)]/[x(2x + y)]

**Solution:**

We have,

(dy/dx) = [y(x + 2y)]/[x(2x + y)]

It is a homogeneous equation,

So, put y = vx (i)

*On differentiating both sides w.r.t x,*

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = [vx(x + 2vx)]/[x(2x + vx]

x(dv/dx) = [vx(x + 2vx)]/[x(2x + vx] – v

x(dv/dx) = (v + 2v^{2 }– 2v – v^{2})/(2 + v)

x(dv/dx) = (v^{2 }– v)/(2 + v)

(2 + v)dv/[v(v – 1)] = (dx/x)

On integrating both sides,

Using partial derivative,

2 + v = A(v – 1) + B(v)

2 + v = v(A + B) – A

On comparing the coefficients,

A = -2

B = 3

-2∫(dv/v) + 3∫dv/(v – 1) = ∫(dx/x)

-2log|v| + 3log|v – 1| = log|x| + log|c|

log|(v – 1)^{3}/v^{2}| = log|xc|

(v – 1)^{3 }= v^{2}|xc|

(y – x)^{3}/x^{3 }= (y/x)^{2}|xc|

(y – x)^{3} = y^{2}x^{2}c

At x = 1, y = 2,

(2 – 1)^{3} = 4 * 1 * c

c = (1/4)

(y – x)^{3} = (1/4)y^{2}x^{2}

### Question 36(vi). (y^{4 }– 2x^{3}y)dx + (x^{4 }– 2xy^{3})dy = 0, y(1) = 0

**Solution:**

We have,

(y^{4 }– 2x^{3}y)dx + (x^{4 }– 2xy^{3})dy = 0

dy/dx = (2x^{3}y – y^{4})/(x^{4 }– 2xy^{3})

It is a homogeneous equation,

So, put y = vx (i)

*On differentiating both sides w.r.t x,*

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (2x^{3}vx – v^{4}x^{4})/(x^{4 }– 2xv^{3}x^{3})

v + x(dv/dx) = (2v – v^{4})/(1 – 2v^{3})

x(dv/dx) = [(2v – v^{4})/(1 – 2v^{3})] – v

x(dv/dx) = (2v – v^{4 }– v + 2v^{4})/(1 – 2v^{3})

x(dv/dx) = (v + v^{4})/(1 – 2v^{3})

On integrating both sides,

∫(dv/v) – ∫(3v^{2})dv/(1 + v^{3}) = log|x| + log|c|

log|v| – log|1 + v^{3}| = log|xc|

log|v/(1 + v^{3})| = log|xc|

At x = 1, y = 1,

1/(1 + 1) = c

c = (1/2)

(yx^{2})/(x^{3 }+ y^{3}) = (1/2)x

### Question 36(vii). x(x^{2 }+ 3y^{2})dx + y(y^{2 }+ 3x^{2})dy = 0, y(1) = 1

**Solution:**

We have,

x(x^{2 }+ 3y^{2})dx + y(y^{2 }+ 3x^{2})dy = 0

dy/dx = -[x(x^{2 }+ 3y^{2})/y(y^{2 }+ 3x^{2})]

It is a homogeneous equation,

So, put y = vx (i)

*On differentiating both sides w.r.t x,*

dy/dx = v + x(dv/dx)

So,

x(dv/dx) = -(1 + 3v^{2 }+ v^{4 }+ 3v^{2})/v(v^{2 }+ 3)

[(v^{3 }+ 3v)/(1 + 6v^{2 }+ v^{4})]dv = -(dx/x)

Multiply both sides with 4 and integrating,

log|v^{4 }+ 6v^{2 }+ 1| = -log|x|^{4 }+ log|c|

|v^{4 }+ 6v^{2 }+ 1| = |c/x^{4}|

(y^{4 }+ 6x^{2}y^{2 }+ x^{4}) = c

At y = 1, x = 1

(1 + 6 + 1) = c

c = 8

(y^{4 }+ 6x^{2}y^{2 }+ x^{4}) = 8

### Question 36(viii). {xsin^{2}(y/x) – y}dx + xdy = 0, y(1) = π/4

**Solution:**

We have,

{xsin

^{2}(y/x) – y}dx + xdy = 0dy/dx = [y – xsin

^{2}(y/x)]/xdy/dx = (y/x) – sin

^{2}(y/x)It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = v – sin

^{2}(v)x(dv/dx) = v – sin

^{2}(v) – vx(dv/dx) = -sin

^{2}(v)-cosec

^{2}(v)dv = (dx/x)On integrating both sides,

-∫cosec

^{2}(v) = ∫(dx/x)cot(v) = log|x| + log|c|

cot(y/x) = log|xc|

At x = 1, y = π/4

cot(π/4) = log|c|

log|c| = 1

c = e

cot(y/x) = log|ex|

### Question 36(ix). x(dy/dx) – y + xsin(y/x) = 0, y(2) = π

**Solution:**

We have,

x(dy/dx) – y + xsin(y/x) = 0

x(dy/dx) = y – xsin(y/x)

(dy/dx) = (y/x) – sin(y/x)

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = v – sin(v)

x(dv/dx) = v – sin(v) – v

x(dv/dx) = -sin(v)

-cosec(v)dv = (dx/x)

On integrating both sides,

∫cosec(v) = -∫(dx/x)

log|cosec(v) – cot(v)| = -log|x| + log|c|

log|cosec(v) – cot(v)| = -log|x| + log|c|

log|cosec(y/x) – cot(y/x)| = -log|x| + log|c|

At x = 2, y = π

|cosec(π/2) – cot(π/2)| = -log|2| + log|c|

log|c| = 0

log|cosec(y/x) – cot(y/x)| = -log|x|

### Question 37. xcos(y/x)(dy/dx) = ycos(y/x) + x, When x = 1, y = π/4

**Solution:**

We have,

xcos(y/x)(dy/dx) = ycos(y/x) + x

(dy/dx) = (y/x) + [1/cos(y/x)]

So, put y = vx (i)

*On differentiating both sides w.r.t x,*

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = v + 1/cosv

x(dv/dx) = v + 1/cosv – v

x(dv/dx) = 1/cosv

cosvdv = (dx/x)

On integrating both sides,

∫cosvdv = ∫(dx/x)

sin(v) = log|x| + log|c|

sin(y/x) = log|x| + log|c|

At x = 1, y = π/4

1/√2 = 0 + log|c|

log|c| = (1/√2)

sin(y/x) = log|x| + (1/√2)

### Question 38. (x – y)(dy/dx) = (x + 2y), when x = 1,y = 0

**Solution:**

We have,

(x – y)(dy/dx) = (x + 2y)

(dy/dx) = (x + 2y)/(x – y)

It is a homogeneous equation,

So, put y = vx (i)

*On differentiating both sides w.r.t x,*

dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x + 2vx)/(x – vx)

v + x(dv/dx) = (1 + 2v)/(1 – v)

x(dv/dx) = [(1 + 2v)/(1 – v)] – v

x(dv/dx) = (1 + 2v – v + v^{2})/(1 – v)

(1 – v)dv/(1 + v + v^{2}) = (dx/x)

On integrating both sides,

At x = 1, y = 0

√3tan^{-1}|1/√3| – (1/2)log|1| = c

c = √3(π/6)

c = (π/2√3)

### Question 39. (dy/dx) = xy/(x^{2 }+ y^{2})

**Solution:**

We have,

(dy/dx) = xy/(x

^{2 }+ y^{2})It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = xvx/(x

^{2 }+ v^{2}x^{2})v + x(dv/dx) = v/(1 + v

^{2})x(dv/dx) = [v/(1 + v

^{2})] – vx(dv/dx) = (v – v – v

^{3})/(1 + v^{2})[-(1/v

^{3}) – (1/v)]dv = (dx/x)On integrating both sides,

-∫dv/v

^{3 }– ∫dv/v = ∫(dx/x)(1/2v

^{2}) – log|v| = log|x| + c(x

^{2}/2y^{2}) = log|vx| + c(x

^{2}/2y^{2}) = log|(y/x)x| + c(x

^{2}/2y^{2}) = log|y| + cAt x = 0, y = 1

c = 0

(x

Solve the following differential equations:^{2}/2y^{2}) = log|y|

Question 1. dy/dx + 2y = e^{3x}Solution:

We have,

dy/dx + 2y = e^{3x}………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2, Q = e^{3x}

So, I.F = e^{∫Pdx}

= e^{∫2dx}

= e^{2x}

The solution of differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^{2x}) = ∫e^{3x}.e^{2x}dx + c

y(e^{2x}) = (1/5)e^{5x }+ c

y = (e^{3x}/5) + ce^{-2x}

Hence, this is the required solution.

Question 2. 4(dy/dx) + 8y = 5e^{-3x}Solution:

We have,

4(dy/dx) + 8y = 5e^{-3x}

(dy/dx) + 2y = (5/4)e^{-3x}………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2, Q = (5/4)e^{-3x}

So, I.F = e^{∫Pdx}

= e^{∫2dx}

= e^{2x}

The solution of differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^{2x}) = (5/4)∫e^{-3x}.e^{2x}dx + c

y(e^{2x}) = (5/4)∫e^{-x}dx + c

y = -(5/4)e^{-3x }+ ce^{-2x}

This is the required solution.

Question 3. (dy/dx) + 2y = 6e^{x}Solution:

We have,

(dy/dx) + 2y = 6e^{x}………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2, Q = 6e^{x}

So, I.F = e^{∫Pdx}

= e^{∫2dx}

= e^{2x}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^{2x}) = ∫6e^{x}.e^{2x}dx + c

y(e^{2x}) = 6∫e^{3x}dx + c

y(e^{2x}) = 2e^{3x }+ c

y = 2e^{x }+ ce^{-2x}

This is the required solution.Question 4.(dy/dx) + y = e^{-2x}Solution:

We have,

(dy/dx) + y = e^{-2x}………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1, Q = e^{-2x}

So, I.F = e^{∫Pdx}

= e^{∫dx}

= e^{x}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^{x}) = ∫e^{-2x}.e^{x}dx + c

y(e^{x}) = ∫e^{-x}dx + c

y(e^{x}) = -e^{-x }+ c

y = -e^{-2x }+ ce^{-x}

This is the required solution.

Question 5. x(dy/dx) = x + ySolution:

We have,

x(dy/dx) = x + y

(dy/dx) = 1 + (y/x)

(dy/dx) – (y/x) = 1 ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = (-1/x), Q = 1

So, I.F = e^{∫Pdx}

= e^{-∫(dx/x)}

= e^{-log(x)}

= e^{log(1/x)}

= (1/x)

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(1/x) = ∫(1/x)dx + c

(y/x) = log|x| + c

y = xlog|x| + cx

This is the required solution.

Question 6. (dy/dx) + 2y = 4xSolution:

We have,

(dy/dx) + 2y = 4x ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2, Q = 4x

So,

I.F = e^{∫Pdx}

= e^{∫2dx}

= e^{2x}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^{2x}) = ∫4x.e^{2x}dx + c

y(e^{2x}) = 4x∫e^{2x}dx – 4∫{(dx/dx)∫e^{2x}dx}dx + c

y(e^{2x}) = 2xe^{2x }– 2∫e^{2x}dx + c

y(e^{2x}) = 2xe^{2x }– e^{2x }+ c

y = (2x – 1) + ce^{-2x}

This is the required solution.

Question 7. x(dy/dx) + y = xe^{x}Solution:

We have,

x(dy/dx) + y = xe^{x}

(dy/dx) + (y/x) = e^{x}………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = (1/x), Q = e^{x}

So,

I.F = e^{∫Pdx}

= e^{∫(dx/x)}

= e^{log(x)}

= x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(x) = ∫x.e^{x}dx + c

xy = x∫e^{x}dx – {(dx/dx)∫e^{x}dx}dx + c

xy = xe^{x }– ∫e^{x}dx + c

xy = xe^{x }– e^{x }+ c

This is the required solution.

Question 8. (dy/dx) + [4x/(x^{2 }+ 1)]y = -1/(x^{2 }+ 1)^{2}Solution:

We have,

(dy/dx) + [4x/(x^{2 }+ 1)]y = -1/(x^{2 }+ 1)^{2}………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = [4x/(x^{2 }+ 1)], Q = -1/(x^{2 }+ 1)^{2}

So,

I.F = e^{∫Pdx}

= (x^{2}+1)^{2}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(x^{2 }+ 1)^{2 }= ∫-[1/(x^{2 }+ 1)^{2}](x^{2 }+ 1)^{2}dx + c

y(x^{2 }+ 1)^{2 }= -∫dx + c

y(x^{2 }+ 1)^{2 }= -x + c

y = -x/(x^{2 }+ 1)^{2 }+ c/(x^{2 }+ 1)^{2}

This is the required solution.

Question 9. x(dy/dx) + y = xlogxSolution:

We have,

x(dy/dx) + y = xlogx

(dy/dx) + (y/x) = logx ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = (1/x), Q = logx

So,

I.F = e^{∫Pdx}

= e^{∫(dx/x)}

= e^{log(x)}

= x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(x) = ∫x.logxdx + c

xy = logx∫xdx – {(d/dx)logx∫xdx}dx + c

xy = (x^{2}/2)logx – ∫(1/x)(x^{2}/2) + c

xy = (x^{2}/2)logx – (1/2)∫xdx + c

xy = (x^{2}/2)logx – (x^{2}/4) + c

y = (x/2)logx – (x/4) + (c/x)

This is the required solution.

Question 10. x(dy/dx) – y = (x – 1)e^{x}Solution:

We have,

x(dy/dx) – y = (x – 1)e^{x}

(dy/dx) – (y/x) = [(x – 1)/x]e^{x}………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -(1/x), Q = [(x – 1)/x]e^{x}

So,

I.F = e^{∫Pdx}

= e^{-∫(dx/x)}

= e^{-log(x)}

= e^{log(1/x)}

= 1/x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

(y/x) = ∫[(1/x) – (1/x^{2})]e^{x }+ c

Since, ∫[f(x) + f'(x)]e^{x}dx = f(x)e^{x }+ c

(y/x) = (e^{x}/x) + c

y = e^{x }+ xc

This is the required solution.

Question 11. (dy/dx) + y/x = x^{3}Solution:

We have,

(dy/dx) + y/x = x^{3}………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/x, Q = x^{3}

So, I.F = e^{∫Pdx}

= e^{∫dx/x}

= e^{logx}

= x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

yx = ∫x^{3}.xdx + c

yx = ∫x^{4}dx + c

yx = (x^{5}/5) + c

y = (x^{4}/5) + c/x

This is the required solution.

Question 12. (dy/dx) + y = sinxSolution:

We have,

(dy/dx) + y = sinx ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1, Q = sinx

So, I.F = e^{∫Pdx}

= e^{∫dx}

= e^{x}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^{x}) = ∫sinx.e^{x}dx + c

Let, I = ∫sinx.e^{x}dx

I = e^{x}∫sinxdx – ∫{(d/dx)e^{x}∫sinxdx}dx

I = -e^{x}cos + ∫e^{x}cosxdx

I = -e^{x}cosx + e^{x}∫cosxdx – ∫{(d/dx)e^{x}∫cosxdx}dx

I = -e^{x}cosx + e^{x}sinx – ∫e^{x}sinxdx

2I = e^{x}(sinx – cosx)

I = (e^{x}/2)(sinx – cosx)

y(e^{x}) = (e^{x}/2)(sinx – cosx) + c

y = (1/2)(sinx – cosx) + ce^{-x}

This is the required solution.

Question 13. (dy/dx) + y = cosxSolution:

We have,

(dy/dx) + y = cosx ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1, Q = cosx

So, I.F = e^{∫Pdx}

= e^{∫dx}

= e^{x}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^{x}) = ∫cosx.e^{x}dx + c

Let, I = ∫cosx.e^{x}dx

I = e^{x}∫cosxdx – ∫{(d/dx)e^{x}∫cosxdx}dx

I = e^{x}sinx – ∫e^{x}sinxdx

I = e^{x}sinx – e^{x}∫sinxdx + ∫{(d/dx)e^{x}∫sinxdx}dx

I = e^{x}sinx + e^{x}cosx – ∫e^{x}cosxdx

2I = e^{x}(cosx + sinx)

I = (e^{x}/2)(cosx + sinx)

y(e^{x}) = (e^{x}/2)(cosx + sinx) + c

y = (1/2)(cosx + sinx) + ce^{-x}

This is the required solution.

Question 14. (dy/dx) + 2y = sinxSolution:

We have,

(dy/dx) + 2y = sinx ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2, Q = sinx

So, I.F = e^{∫Pdx}

= e^{∫2dx}

= e^{2x}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^{2x}) = ∫sinx.e^{2x}dx + c

Let, I = ∫sinx.e^{2x}dx

I = e^{2x}∫sinxdx – {(d/dx)e^{2x}∫sinxdx}dx

I = -e^{2x}cosx + 2∫e^{2x}cosdx

I = -e^{2x}cosx + 2e^{2x}∫cosxdx – 2{(d/dx)e^{2x}∫cosxdx}dx

I = -e^{2x}cosx + 2e^{2x}sinx – 4∫e^{2x}sinxdx

5I = e^{2x}(2sinx – cosx)

I = (e^{2x}/5)(2sinx – cosx)

y(e^{2x}) = (e^{2x}/5)(2sinx – cosx) + c

y = (1/5)(2sinx – cosx) + ce^{-2x}

This is the required solution.

Question 15. (dy/dx) – ytanx = -2sinxSolution:

We have,

(dy/dx) – ytanx = -2sinx ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -tanx, Q = sinx

So,

I.F = e^{∫Pdx}

= e^{∫-tanxdx}

= e^{-log|secx|}

= 1/secx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(1/secx) = -2∫sinx.(1/secx)dx + c

ycosx = -∫2sinx.cosxdx + c

ycosx = -∫sin2xdx + c

ycosx = (cos2x/2) + c

y = (cos2x/cosx) + (c/cosx)

This is the required solution.

Question 16. (1 + x^{2})(dy/dx) + y = tan^{-1}xSolution:

We have,

(1 + x^{2})(dy/dx) + y = tan^{-1}x

(dy/dx) + [y/(1 + x^{2})] = tan^{-1}x/(1 + x^{2}) ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/(1 + x^{2}), Q = tan^{-1}x/(1 + x^{2})

So,

I.F = e^{∫Pdx}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

Let, tan^{-1}x = z

On differentiating both sides we get,

dx/(1 + x^{2}) = dz

y(e^{z}) = ∫ze^{z}dz + c

y(e^{z}) = z∫e^{z}dz – {(dz/dz)∫e^{z}dz}dz

y(e^{z}) = ze^{z }– ∫ e^{z}dz + c

y(e^{z}) = e^{z}(z – 1) + c

y = (z – 1) + ce^{-z}

This is the required solution.

Question 17. (dy/dx) + ytanx = cosxSolution:

We have,

(dy/dx) + ytanx = cosx ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = tanx, Q = cosx

So, I.F = e^{∫Pdx}

= e^{∫tanxdx}

= e^{log|secx|}

= secx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y.secx = ∫cosx.secxdx + c

y.secx = ∫dx + c

y.secx = x + c

y = xcosx + c.cosx

This is the required solution.

Question 18. (dy/dx) + ycotx = x^{2}cotx + 2xSolution:

We have,

(dy/dx) + ycotx = x^{2}cotx + 2x ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = cotx, Q = x^{2}cotx + 2x

So,

I.F = e^{∫Pdx}

= e^{∫cotxdx}

= e^{log|sinx|}

= sinx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(sinx) = ∫(x^{2}cotx + 2x)sinxdx + c

y(sinx) = ∫(x^{2}cosx + 2xsinx)dx + c

y(sinx) = x^{2}∫cosxdx – {(d/dx)x^{2}∫cosxdx}dx + ∫2xsinxdx + c

y(sinx) = x^{2}sinx – ∫2xsinxdx + ∫2xsinxdx + c

y(sinx) = x^{2}sinxdx + c

This is the required solution.

Question 19. (dy/dx) + ytanx = x^{2}cos^{2}xSolution:

We have,

(dy/dx) + ytanx = x^{2}cos^{2}x ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = tanx, Q = x^{2}cos^{2}x

So,

I.F = e^{∫Pdx}

= e^{∫tanxdx}

= e^{log|secx|}

= secx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(secx) = ∫secx.(x^{2}.cos^{2}x)dx + c

y(secx) = ∫x^{2}.cosxdx + c

y(secx) = x^{2}∫cosxdx – ∫{(d/dx)x^{2}∫cosxdx}dx + c

y(secx) = x^{2}sinx – 2∫xsinxdx + c

y(secx) = x^{2}sinx – 2x∫sinxdx + 2∫{(dx/dx)∫sinxdx}dx + c

y(secx) = x^{2}sinx + 2xcosx – 2∫cosxdx + c

y(secx) = x^{2}sinx + 2xcosx – 2sinx + c

This is the required solution.Question 20.(1 + x^{2})(dy/dx) + y =Solution:

We have,

(1 + x^{2})(dy/dx) + y =

(dy/dx) + [1/(x^{2 }+ 1)]y = /(x^{2 }+ 1) ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/(x^{2 }+ 1), Q = /(x^{2}+ 1)

So,

I.F = e^{∫Pdx}

=

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y() – ∫[/(x^{2 }+ 1)]dx + c

Let, tan^{-1}x = z

On differentiating both sides we get

dx/(1 + x^{2}) = dz

ye^{z }= ∫e^{2z}dz + c

ye^{z }= (e^{2z}/2) + c

y = (e^{z}/z) + c.e^{-z}

y = (1/2)^{ }+ c.

This is the required solution.

Question 21. xdy = (2y + 2x^{4 }+ x^{2})dxSolution:

We have,

xdy = (2y + 2x^{4 }+ x^{2})dx

(dy/dx) = 2(y/x) + 2x^{3 }+ x

(dy/dx) – (y/x) = 2x^{3 }+ x ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -(2/x), Q = 2x^{3 }+ x

So,

I.F = e^{∫Pdx}

= e^{-2∫(dx/x)}

= e^{-2log(x)}

= e^{2log|1/x|}

= 1/x^{2}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(1/x^{2}) = ∫(1/x^{2}).(2x^{3 }+ x)dx + c

(y/x^{2}) = ∫[2x + (1/x)]dx + c

(y/x^{2}) = x^{2 }+ log|x| + c

y = x^{4 }+ x^{2}log|x| + cx^{2}

This is the required solution.

Question 22. (1 + y^{2}) + (x – )(dy/dx) = 0Solution:

We have,

(1 + y^{2}) + (x – )(dy/dx) = 0

(dx/dy) + x/(1 + y^{2}) = /(1 + y^{2}) ………..(i)

The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = 1/(1 + y^{2}), Q = /(1 + y^{2})

So,

I.F = e^{∫Pdy}

=

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

Let, tan^{-1}y = z

On differentiating both sides we have,

dy/(1 + y^{2}) = dz

xe^{z }= ∫e^{2z}dz + c

xez = (e2z/2) + c

x = e^{z}/2 + ce^{-z}

This is the required solution.

Question 23. y^{2}(dx/dy) + x – 1/y = 0Solution:

We have,

y^{2}(dx/dy) + x – 1/y = 0

(dx/dy) + (x/y^{2}) = 1/y^{3}………..(i)

The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = 1/y^{2}, Q = 1/y^{3}

So,

I.F = e^{∫Pdy}

= e^{∫dy/y2}

= e^{-(1/y)}

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

e^{-(1/y)}x = ∫(1/y^{3}).(e^{-1/y})dy + c

Let,-(1/y) = z

Differentiating both sides we have,

(dy/y^{2}) = dz

xe^{z }= -∫ze^{z}dz + c

xe^{z }= -z∫e^{z}dz + ∫{(dz/dz)∫e^{z}dz}dz + c

xe^{z }= -ze^{z }+ ∫e^{z}dz + c

xe^{z }= -ze^{z }+ e^{z }+ c

x = (1 – z) + ce – z

x = [1 + (1/y)] + ce^{1/y}

x = (y + 1)/y + ce^{1/y}

This is the required solution.

Solve the following differential equations:

Question 24. (2x – 10y^{3})(dy/dx) + y = 0Solution:

We have,

(2x – 10y^{3})(dy/dx) + y = 0

(2x – 10y^{3})(dy/dx) = -y

(dx/dy) = -(2x – 10y^{3})/y

(dx/dy) + 2x/y = 10y^{2}………..(i)The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = 2/y, Q = 10y^{2}

So,

I.F = e^{∫Pdy}

= e^{∫(2/y)dy}

= e^{2log|y|}

= y^{2}

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(y^{2}) = ∫(10y^{2})(y^{2})dy + c

xy^{2}= 10(y^{5}/5) + c

x = 2y^{3 }+ cy^{-2}

This is the required solution.

Question 25. (x + tany)dy = sin2ydxSolution:

We have,

(x + tany)dy = sin2ydx

(dx/dy) = (x + tany)/sin2y

(dx/dy) – cosec2y.x = tany/sin2y………..(i)The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = -cosec2y, Q = tany/sin2y

So, I.F = e^{∫Pdy}

= e^{∫-cosec2ydy}

=

=

= √coty

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(√coty) = ∫(tany/sin2y).(√coty)dy + c

Let, tany = z

On differentiating both side we have,

sec^{2}ydx = dz

(x/√tany) = (1/2)∫dz/√z + c

(x/√tany) = (1/2)(2√z) + c

x = (√tany)(√tany) + c(√tany)

x = tany + c(√tany)

This is the required solution.

Question 26. dx + xdy = e^{-y}sec^{2}ydySolution:

We have,

dx + xdy = e^{-y}sec^{2}ydy

(x – e^{-y}sec^{2}y)dy = -dx

(dx/dy) = (e^{-y}sec^{2}y-x)………..(i)

The above equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = 1, Q = e^{-y}sec^{2}y

So, I.F = e^{∫Pdy}

= e^{∫dy}

= e^{y}

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(e^{y}) = ∫e^{-y}sce^{2}ye^{y}dy + c

x(e^{y}) = ∫sec^{2}ydy + c

x(e^{y}) = tany + c

x = (tany + c)e^{-y}

This is the required solution.

Question 27. (dy/dx) = ytanx – 2sinxSolution:

We have,

(dy/dx) = ytanx – 2sinx

(dy/dx) – ytanx = -2sinx………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -tanx, Q = sinx

So,

I.F = e^{∫Pdx}

= e^{∫-tanxdx}

= e^{-log|secx|}

= 1/secx

= cosx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(cosx) = -2∫sinx.(cosx)dx + c

ycosx = -∫2sinx.cosxdx + c

Let, sinx = z

On differentiating both sides we have,

cosxdx = dz

ycosx = -2∫zdz + c

ycosx = -2(z^{2}/2) + c

ycosx = -sin^{2}xdx + c

y = secx(-sin^{2}xdx + c)

This is the required solution.

Question 28. (dy/dx) + ycosx = sinx.cosxSolution:

We have,

(dy/dx) + ycosx = sinx.cosx………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = cosx, Q = sinx.cosx

So,

I.F = e^{∫Pdx}

= e^{∫cosxdx}

= e^{sinx}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^{sinx}) = ∫(e^{sinx})(sinx.cosx)dx + c

Let, sinx = z

Differentiating both sides we get,

cosxdx = dz

y(e^{z}) = ∫ze^{z}dz + c

y(e^{z}) = z∫e^{z}dz – {(dz/dz)∫e^{z}dz}dz

y(e^{z}) = ze^{z }– ∫e^{z}dz + c

y(e^{z}) = e^{z}(z – 1) + c

y = (z – 1) + ce^{-z}

y = (sinx – 1) + ce^{-sinx}

This is the required solution.

Question 29. (1 + x^{2})(dy/dx) – 2xy = (x^{2 }+ 2)(x^{2 }+ 1)Solution:

We have,

(1 + x^{2})(dy/dx) – 2xy = (x^{2 }+ 2)(x^{2 }+ 1)

(dy/dx) – 2xy/(1 + x^{2}) = (x^{2 }+ 2)………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -2x/(1 + x^{2}), Q = (x^{2 }+ 1)

So,

I.F = e^{∫Pdx}

=

= 1/(x^{2}+1)

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y.[1/(x^{2 }+ 1)] = ∫[(x^{2 }+ 2)/(x^{2 }+ 1)]dx + c

y/(x^{2 }+ 1) = ∫[1 + 1/(x^{2 }+ 1)]dx + c

y/(x^{2 }+ 1) = x + tan^{-1}x + c

y = (x^{2 }+ 1)(x + tan^{-1}x + c)

This is the required solution.

Question 30. sinx(dy/dx) + ycosx = 2sin^{2}xcosxSolution:

We have,

sinx(dy/dx) + ycosx = 2sin^{2}xcosx

(dy/dx) + ycotx = 2sinx.cosx………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = cotx, Q = 2sinx.cosx

So,

I.F = e^{∫Pdx}

= e^{∫cotxdx}

= e^{log|sinx|}

= sinx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(sinx) = ∫(2sinx.cosx)sinxdx + c

Let, sinx = z

On differentiating both sides we have,

cosxdx = dz

y.z = 2∫z^{2 }+ c

y.z = (2/3)z^{3 }+ c

y.sinx = (2/3)sin^{3}x + c

This is the required solution.

Question 31. (x^{2 }– 1)(dy/dx) + 2(x + 2)y = 2(x + 1)Solution:

We have,

(x^{2 }– 1)(dy/dx) + 2(x + 2)y = 2(x + 1)

(dy/dx) + 2(x + 2)y/(x^{2 }– 1) = 2(x + 1)/(x^{2}– 1)

(dy/dx) + 2(x + 2)y/(x^{2 }– 1) = 2/(x – 1)………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2(x + 2)/(x^{2 }– 1), Q = 2/(x – 1)

So,

I.F = e^{∫Pdx}

=

= (x – 1)^{3}/(x + 1)

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y.[(x – 1)^{3}/(x + 1)] = ∫[(x – 1)^{3}/(x + 1)[{2/(x – 1)]dx + c

y.[(x – 1)^{3}/(x + 1)] = (x^{2 }– 6x + 8log|x + 1|) + c

This is the required solution.

Question 32. (dy/dx) + (2y/x) = cosxSolution:

We have,

(dy/dx) + (2y/x) = cosx………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2/x, Q = cosx

So,

I.F = e^{∫Pdx}

= e^{∫(2/x)dx}

= e^{2log|x|}

= x^{2}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(x^{2}) = ∫(x^{2}).(cosx)dx + c

x^{2}(y) = x^{2}∫cosxdx – ∫{(d/dx)x^{2}∫cosxdx}dx + c

x^{2}y = x^{2}sinx – 2∫xsinxdx + c

x^{2}y = x^{2}sinx – 2x∫sinxdx + 2∫{(dx/dx)∫sinxdx}dx + c

x^{2}y = x^{2}sinx + 2xcosx – 2∫cosxdx + c

x^{2}y = x^{2}sinx + 2xcosx – 2sinx + c

This is the required solution.

Question 33. (dy/dx) – y = xe^{x}Solution:

We have,

(dy/dx) – y = xe^{x}………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -1, Q = xe^{x}

So,

I.F = e^{∫Pdx}

= e^{∫-dx}

= e^{-x}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^{-x}) = ∫(e^{-x})(xe^{x})dx + c

ye^{-x }= ∫xdx + c

ye^{-x }= (x^{2}/2) + c

y = [(x^{2}/2) + c].e^{x}

This is the required solution.

Question 34. (dy/dx) + 2y = xe^{4x}Solution:

We have,

(dy/dx) + 2y = xe^{4x}………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2, Q = xe^{4x}

So,

I.F = e^{∫Pdx}

= e^{2∫dx}

= e^{2x}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^{2x}) = ∫(e^{2x}).(xe^{4x})dx + c

y(e^{2x}) = ∫xe^{6x}dx + c

y(e^{2x}) = x∫e^{6x}dx – ∫{(dx/dx)∫e^{6x}dx}dx + c

e^{2x}y = (xe^{6x})/6 – ∫(e^{6x}/6)dx + c

e^{2x}y = (xe^{6x})/6 – e^{6x}/36 + c

y = (xe^{4x})/6 – e^{4x}/36 + ce^{-2x}

This is the required solution.

Question 35. (x + 2y^{2})(dy/dx) = y, given that when x = 2, y = 1Solution:

We have,

(x + 2y^{2})(dy/dx) = y

(dx/dy) = (x + 2y^{2})/y

(dx/dy) – (x/y) = 2y………..(i)The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = 1/y, Q = 2y

So,

I.F = e^{∫Pdy}

= e^{∫-dy/y}

= e^{-log|y|}

= 1/y

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(1/y) = ∫(1/y)(2y)dy + c

(x/y) = 2∫dy + c

(x/y) = 2y + c

x = 2y^{2 }+ cy

Given that when x = 2, y = 1

2 = 2 + c

c = 0

x = 2y^{2}

This is the required solution.

Question 36(i). Find one-parameter families of solution curves of the following differential equation (dy/dx) + 3y = e^{mx}, m is a given real numberSolution:

We have,

(dy/dx) + 3y = e^{mx}………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 3, Q = e^{mx}

So,

I.F = e^{∫Pdx}

= e^{∫3dx}

= e^{3x}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^{3x}) = ∫(e^{3x}).(e^{mx})dx + c

y(e^{3x}) = ∫e^{(3+m)x}dx + c

y(e^{3x}) = e^{(m+3)x}/(m + 3) + c

y = e^{mx}/(m + 3) + c

This is the required solution.

Question 36(ii). Find one-parameter families of solution curves of the following differential equation (dy/dx) – y = cos2xSolution:

We have,

(dy/dx) – y = cos2x………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -1, Q = cos2x

So,

I.F = e^{∫Pdx}

= e^{-∫dx}

= e^{-x}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^{-x}) = ∫(e^{-x}).(cos2x)dx + c

y(e^{-x}) = ∫e^{-x}cos2xdx + c

Let,

A = ∫e^{-x}cos2xdx

= e^{-x}∫cos2xdx – {(d/dx)e^{-x}∫cos2xdx}dx

= (e^{-x}/2)sin2x + ∫(e^{-x}/2)sin2xdx

=

(5/4)A = (e^{-x}/2)(2sin2x – cos2x)

This is the required solution.

Question 36(iii). Find one-parameter families of solution curves of the following differential equation x(dy/dx) – y = (x + 1)e^{-x}Solution:

We have,

x(dy/dx) – y = (x + 1)e^{-x}

(dy/dx) – y/x = [(x + 1)/x]e^{-x }………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -1/x, Q = [(x + 1)/x]e^{-x}

So,

I.F = e^{∫Pdx}

= e^{-∫dx/x}

= e^{-log|x|}

= 1/x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(1/x) = ∫[(x+1)/x]e^{-x}(1/x)dx + c

y/x = ∫[1/x+1/x^{2}]e^{-x}dx + c

Let, (1/x)e^{-x }= z

On differentiating both sides we have

-[1/x + 1/x2]e^{-x}dx = dz

y/x = -∫dz + c

y/x = -z + c

y/x = -(e^{-x}/x) + c

y = -e^{-x }+ cx

This is the required solution.

Question 36(iv). Find one-parameter families of solution curves of the following differential equation x(dy/dx) + y = x^{4}Solution:

We have,

x(dy/dx) + y = x^{4 }

(dy/dx) + y/x = x^{3}………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/x, Q = x^{3}

So,

I.F = e^{∫Pdx}

= e^{∫dx/x}

= e^{log|x|}

= x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(x) = ∫(x)(x^{3})dx + c

xy = ∫x^{4 }+ c

xy = (x^{5}/5) + c

y = (x^{4}/5) + c/x

This is the required solution.

Question 36(v). Find one-parameter families of solution curves of the following differential equation (xlogx)(dy/dx) + y = logxSolution:

We have,

(xlogx)(dy/dx) + y = logx

(dy/dx) + y/xlogx = 1/x………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/xlogx, Q = 1/x

So,

I.F = e^{∫Pdx}

= e^{∫dx/xlogx}

Let, logx = z

On differentiating both sides we have

dx/x = dz

= e^{∫dz/z}

= e^{log|z|}

= z

=l ogx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(logx) = ∫(1/x)(logx)dx + c

y(logx) = ∫zdz + c (Let, logx=z and differentiating both sides)

y(logx) = (z^{2}/2) + c

y(logx) = (logx)^{2}/2 + c

y = logx/2 + c/logx

This is the required solution.

Question 36(vi). Find one-parameter families of solution curves of the following differential equation (dy/dx) – 2xy/(1 + x^{2}) = x^{2 }+ 2Solution:

We have,

(dy/dx) – 2xy/(1 + x^{2}) = x^{2 }+ 2………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -2x/(1 + x^{2}), Q = x^{2 }+ 2

So,

I.F = e^{∫Pdx}

= e^{-∫2xdx/(1+x2)}

= e^{-log|1+x2|}

= 1/(1+x^{2})

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y[1/(1 + x^{2})] = ∫[1/(1 + x^{2})](x^{2 }+ 2)dx + c

y/(1 + x^{2}) = ∫[(x^{2 }+ 2)/(x^{2 }+ 1)]dx + c

y/(1 + x^{2}) = ∫dx + ∫dx/(x^{2 }+ 1) + c

y/(x^{2 }+ 1) = x + tan^{-1}x + c

y = (x + tan^{-1}x + c)(x^{2 }+ 1)

This is the required solution.

Question 36(vii). Find one-parameter families of solution curves of the following differential equation (dy/dx) + ycosx = e^{sinx}cosxSolution:

We have,

(dy/dx) + ycosx = e^{sinx}cosx………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = cosx, Q = e^{sinx}cosx

So,

I.F = e^{∫Pdx}

= e∫cosxdx

= e^{sinx}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y[(e^{sinx}) = ∫(e^{sinx})(e^{sinx}cosx)dx + c

Let, sinx = z

ON differentiating both sides we have,

cosxdx = dz

ye^{z }= ∫e^{2z}dz + c

ye^{z }= (e^{2z}/2) + c

y = ez/2 + ce – z

y = (e^{sinx}/2) + ce^{-sinx}

This is the required solution.

Question 36(viii). Find one-parameter families of solution curves of the following differential equation (x + y)(dy/dx) = 1Solution:

We have,

(x + y)(dy/dx) = 1

(dy/dx) = 1/(x + y)

(dx/dy) = (x + y)

(dx/dy) – x = y………..(i)The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = -1, Q = y

So,

I.F = e^{∫Pdy}

= e^{-∫dy}

= e^{-y}

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(e^{-y}) = ∫(e^{-y})(y)dy + c

xe^{-y }= y∫e^{-y}dy – ∫{(dy/dy)∫e^{-y}dy}dy + c

xe^{-y }= -ye^{-y }+ ∫e^{-y }+ c

xe^{-y }= -ye^{-y }– e^{-y }+ c

xe^{-y }+ ye^{-y }+ e^{-y }= c

e^{-y}(x + y + 1) = c

(x + y + 1) = ce^{y}

This is the required solution.

Question 36(ix). Find one-parameter families of solution curves of the following differential equation cos^{2}x(dy/dx) = (tanx – y)Solution:

We have,

cos^{2}x(dy/dx) = (tanx – y)

(dy/dx) = (tanx – y)/cos^{2}x

(dy/dx) = tanx.sec^{2}x – ysec^{2}x

(dy/dx) + ysec^{2}x = tanx.sec^{2}x………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = sec^{2}x, Q = tanx.sec^{2}x

So,

I.F = e∫Pdx

= e^{tanx}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^{tanx}) = ∫(e^{tanx})(tanx.sec^{2}x)dx + c

Let, tanx = z

On differentiating both sides we have,

sec^{2}xdx = dz

y(e^{z}) = ∫ze^{z}dz + c

y(e^{z}) = z∫e^{z}dz – ∫{(dz/dz)∫e^{z}dz}dz

y(e^{z}) = ze^{z }– ∫e^{z}dz + c

y(e^{z}) = ze^{z }– e^{z }+ c

y = (z – 1) + c.e^{-z}

y = (tanx – 1) + c.e^{-tanx}

This is the required solution.

Question 36(x). Find one-parameter families of solution curves of the following differential equation e^{-y}sec^{2}ydy = dx + xdySolution:

We have,

dx + xdy = e^{-y}sec^{2}ydy

(x – e^{-y}sec^{2}y)dy = -dx

(dx/dy) = (e^{-y}sec^{2}y – x)

(dx/dy) + x = e^{-y}sec^{2}y………..(i)The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = 1, Q = e^{-y}sec^{2}y

So, I.F = e^{∫Pdy}

= e^{∫dy}

= e^{y}

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(e^{y}) = ∫e^{-y}sce^{2}ye^{y}dy + c

x(e^{y}) = ∫sec^{2}ydy + c

x(e^{y}) =tany + c

x = (tan y + c)e^{-y}

This is the required solution.

Question 36(xi). Find one-parameter families of solution curves of the following differential equation (xlogx)(dy/dx) + y = 2logxSolution:

We have,

(xlogx)(dy/dx) + y = 2logx

(dy/dx) + y/xlogx = 2/x………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/xlogx, Q = 2/x

So,

I.F = e^{∫Pdx}

= e^{∫dx/xlogx}

Let, logx = z

On differentiating both sides we have

dx/x = dz

= e^{∫dz/z}

= e^{log|z|}

= z

= logx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(logx) = ∫(2/x)(logx)dx + c

y(logx) = 2∫zdz + c (Let, logx = z and differentiating both sides)

y(logx) = 2(z^{2}/2) + c

y(logx) = (logx)^{2 }+ c

y = logx + c/logx

This is the required solution.

Question 36(xii). Find one-parameter families of solution curves of the following differential equation x(dy/dx) + 2y = x^{2}logxSolution:

We have,

x(dy/dx) + 2y = x^{2}logx

(dy/dx) + 2y/x = xlogx………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2/x, Q = xlogx

So,

I.F = e∫Pdx

= e^{2∫dx/x}

= e^{2logx}

= x^{2}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(x^{2}) = ∫(x^{2})(xlogx)dx + c

x^{2}y = ∫x^{3}logxdx + c

x^{2}y = logx∫x^{3}dx + ∫{(d/dx)logx∫x^{3}dx}dx + c

x^{2}y = (1/4)x^{4}logx – (1/4)∫x^{3}dx + c

x^{2}y = (1/4)x^{4}logx – (1/16)x^{4 }+ c

y = (x^{2 }/16)(4logx – 1) + c/x^{2}

This is the required solution.

Question 37. Solve the following using the initial value problem:-

(i). y’ + y = e^{x}, y(0) = (1/2)Solution:

We have,

y’ + y = e^{x}

dy/dx + y = e^{x}………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1, Q = e^{x}

So, I.F = e^{∫Pdx}

= e^{∫dx}

= e^{x}

The solution of differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^{x}) = ∫e^{x}.e^{x}dx + c

y(e^{x}) = (1/2)e^{2x }+ c

At t = 0, y = (1/2)

(1/2)e^{0 }= (1/2)e^{0 }+ c

c = 0

y(e^{x}) = (1/2)e^{2x}

y = (e^{x}/2)

This is the required solution.

(ii). x(dy/dx) – y = logx, y(1) = 0Solution:

We have,

x(dy/dx) – y = logx

(dy/dx) – y/x = logx/x………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -1/x, Q = logx/x

So,

I.F = e^{∫Pdx}

= e^{-∫dx/x}

= e^{-log|x|}

= 1/x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(1/x) = ∫(1/x)(logx/x)dx + c

(y/x) = ∫(logx/x^{2})dx + c

(y/x) = logx∫(dx/x^{2}) – ∫{(d/dx)logx∫(dx/x^{2})}dx + c

(y/x) = -(logx/x) + ∫(dx/x^{2}) + c

(y/x) = -(logx/x) – (1/x) + c

At x = 1, y = 0

0 = -0 – 1 + c

c = 1

(y/x) = -(logx/x) – (1/x) + 1

y = x – 1 – logx

This is the required solution.

(iii). (dy/dx) + 2y = e^{-2x}sinx, y(0) = 0Solution:

We have,

(dy/dx) + 2y = e^{-2x}sinx………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2, Q = e^{-2x}sinx

So,

I.F = e^{∫Pdx}

= e^{∫2dx}

= e^{2x}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^{2x}) = ∫e^{-2x}sinx.(e^{2x})dx + c

y(e^{2x}) = ∫sinxdx + c

y(e^{2x}) = -cosx + c

At x = 0, y = 0

0 = -1 + c

c = 1

y(e^{2x}) = 1 – cosx

This is the required solution.

(iv). x(dy/dx) – y = (x + 1)e^{-x}, y(1) = 0Solution:

We have,

x(dy/dx) – y = (x + 1)e^{-x}

(dy/dx) – (y/x) = [(x + 1)/x]e^{-x}………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -(1/x), Q = [(x + 1)/x]e^{-x}

So,

I.F = e^{∫Pdx}

= e^{-∫(dx/x)}

= e^{-log(x)}

= e^{log(1/x)}

= 1/x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

(y/x) = ∫[(1/x) + (1/x^{2})]e^{-x }+ c

Since, -∫[f(x) + f'(x)]e^{-x}dx = f(x)e^{-x }+ c

(y/x) = -e^{-x}/x + c

At x = 1, y = 0

0 = -e^{-1 }+ c

c = e^{-1}

(y/x) = -e^{-x}/x + e^{-1}

y = xe^{-1 }– e^{-x}

This is the required solution.

(v). (1 + y^{2})dx + (x – )dx = 0, y(0) = 0Solution:

We have,

(1 + y^{2})(dx/dy) + x =

(dy/dx) + [1/(y^{2 }+ 1)]x = /(y^{2 }+ 1)………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Px = Q

Where, P = 1/(y^{2 }+ 1), Q = /(y^{2 }+ 1)

So,

I.F = e^{∫Pdy}

= e^{tan-1y}

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x() = ∫[/(y^{2 }+ 1)]dx + c

x() = ∫dy/(1 + y^{2}) + c

x(e^{tan-1y}) = tan^{-1}y + c

At x = 0, y = 0

0*e^{0 }= 0 + c

c = 0

x() = tan^{-1}y

This is the required solution.

(vi). (dy/dx) + ytanx = x^{2}tanx + 2x, y(0) = 1Solution:

We have,

(dy/dx) + ytanx = x^{2}tanx + 2x………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = tanx, Q = x^{2}tanx+2x

So,

I.F = e^{∫Pdx}

= e^{∫tanxdx}

= e^{log|secx|}

= secx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(secx) = ∫(x^{2}tanx + 2x)secxdx + c

y(secx) = ∫(x^{2}tanxsecx + 2xsecx)dx + c

y(secx) = ∫x^{2}tanxsecxdx + 2∫xsecxdx + c

y(secx) = ∫x^{2}tanxsecxdx + 2secx∫xdx – 2∫{(d/dx)secx∫xdx}dx + c

y(secx) = ∫x^{2}tanxsecxdx + x^{2}.secx – ∫x^{2}tanxsecxdx + c

y(secx) = x^{2},(secx)+c

At, x = 0, y = 1

1 = 0 + c

c = 1

y = x^{2 }+ cosx

This is the required solution.

(vii). x(dy/dx) + y = xcosx + sinx, y(π/2) = 1Solution:

We have,

x(dy/dx) + y = xcosx + sinx

(dy/dx) + (y/x) = cosx + sinx/x………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/x, Q = cosx + sinx/x

So,

I.F = e^{∫Pdx}

= e^{∫dx/x}

= e^{log|x|}

= x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(x) = ∫(cosx + sinx/x)(x)xdx + c

y(x) = ∫xcosxdx + ∫sinxdx + c

xy = x∫cosxdx – ∫{(dx/dx)∫cosxdx}dx – cosx + c

xy = xsinx – ∫sinxdx – cosx + c

xy = xsinx + cosx – cosx + c

xy = xsinx + c

At x = π/2, y = 1

π/2 = π/2sin(π/2) + c

c = 0

y = sinx

This is the required solution.

(viii). (dy/dx) + ycotx = 4xcosecx, y(π/2) = 0Solution:

We have,

(dy/dx) + ycotx = 4xcosecx………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = cotx, Q = 4xcosecx

So,

I.F = e^{∫Pdx}

= e^{∫cotx.dx}

= e^{log|sinx|}

= sinx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(sinx) = 4∫(xcosecx)(sinx)xdx + c

y(sinx) = 4∫xdx + c

y(sinx) = 4(x^{2}/2) + c

y(sinx) = 2x^{2 }+ c

At x = π/2, y = 0,

0 = 2(π/2)^{2 }+ c

c = -π^{2}/2

y(sinx) = 2x^{2 }– π^{2}/2

This is the required solution.

(ix). (dy/dx) + 2ytanx = sinx, y = 0 when x = π/3Solution:

We have,

(dy/dx) + 2ytanx = sinx………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2tanx, Q = sinx

So, I.F = e^{∫Pdx}

= e^{∫2tanxdx}

= e^{2log|secx|}

= sec^{2}x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y.sec^{2}x = ∫sinx.sec^{2}xdx + c

y.sec^{2}x = ∫tanx.secxdx + c

y.sec^{2}x = secx+c

At x = π/3, y = 0,

0 = sec^{2}(π/3) + c

c = -2

y.sec^{2}x = secx – 2

y = cosx – 2cos^{2}x

This is the required solution.

(x). (dy/dx) – 3ycotx = sin2x, y = 2 when x = π/2Solution:

We have,

(dy/dx) – 3ycotx = sin2x………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -3cotx, Q = sin2x

So, I.F = e^{∫Pdx}

= e^{∫-3cotxdx}

= e^{-3log|sinx|}

= cosec^{3}x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y.(cosec^{3}x) = 2∫(cosec^{3}x).(sin2x)dx + c

y.(cosec^{3}x) = 2∫cotx.cosecxdx + c

y.(cosec^{3}x) = -2cosesx +c

y = -2sin^{2}x + c.sin^{3}x

At x = π/2, y = 2.

2 = -2sin^{2}(π/2) + c.sin^{3}(π/2)

c = 4

y = c.sin^{3}x – 2sin^{2}x

This is the required solution.

(xi). (dy/dx) + ycotx = 2cosx, y(π/2) = 0Solution:

We have,

(dy/dx) + ycotx = 2cosx………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = cotx, Q = 2cosx

So, I.F = e^{∫Pdx}

= e^{∫cotxdx}

= e^{log|sinx|}

= sinx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y.(sinx) = ∫(sinx).(2cosx)dx + c

y.(sinx) = 2∫sinx.cosxdx + c

y.(sinx) = ∫sin2xdx + c

y.(sinx) = -(cos2x/2) + c

At x = π/2, y = 0

0 = -cos(π)/2 + c

c = -(1/2)

y.(sinx) = -(cos2x/2) – (1/2)

2y(sinx) = -(1 + cos2x)

2y(sinx) = -2cos^{2}x

y = -cotx.cosx

This is the required solution.

(xii). dy = cosx(2 – ycosecx)dx,Solution:

We have,

dy = cosx(2 – ycosecx)dx

(dy/dx) = -ycotx + 2cosx

(dy/dx) + ycotx = 2cosx………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = cotx, Q = 2cosx

So,

I.F = e^{∫Pdx}

= e^{∫cotxdx}

= e^{log|sinx|}

= sinx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(sinx) = 2∫cosx.(sinx)dx + c

ysinx = ∫2cosx.sinxdx + c

ysinx = ∫sin2x + c

ysinx = -(cos2x/2) + c

This is the required solution.

Question 38. x(dy/dx) + 2y = x^{2}Solution:

We have,

x(dy/dx) + 2y = x^{2}

(dy/dx) + 2y/x = x………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2/x, Q = x

So, I.F = e^{∫Pdx}

= e^{2∫dx/x}

= e^{2logx}

= x^{2}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

yx^{2 }= ∫x2.xdx + c

yx^{2 }= ∫x^{3}dx + c

x^{2}y = (x^{4}/4) + c

y = (x1/4) + c.x^{-2}

This is the required solution.

Question 39. (dy/dx) – y = cosxSolution:

We have,

(dy/dx) – y = cosx………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -1, Q = cosx

So, I.F = e^{∫Pdx}

= e^{-∫dx}

= e^{-x}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^{-x}) = ∫cosx.e^{-x}dx + c

Let, I = ∫cosx.e^{-x}dx

I = e^{-x}∫cosxdx – ∫{(d/dx)e^{-x}∫cosxdx}dx

I = e^{-x}sinx + ∫e^{-x}sinxdx

I = e^{-x}sinx + e^{-x}∫sinxdx-∫{(d/dx)e^{-x}∫sinxdx}dx

I = e^{-x}sinx – e^{-x}cosx-∫e^{-x}cosxdx

2I = e^{-x}(sinx – cosx)

I = (e^{-x}/2)(sinx – cosx)

y(e^{-x}) = (e^{-x}/2)(sinx – cosx) + c

y = (1/2)(sinx-cosx) + ce^{x}

This is the required solution.

Question 40. (y + 3x^{2})(dx/dy) = xSolution:

We have,

(y + 3x^{2})(dx/dy) = x

(dy/dx) = (y + 3x^{2})/x

(dy/dx) – y/x = 3x………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -1/x, Q = 3x

So, I.F = e^{∫Pdx}

= e^{-∫dx/x}

= e^{-logx}

= 1/x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(1/x) = 3∫x.(1/x)dx + c

y(1/x) = 3∫dx + c

y/x = 3x + c

This is the required solution.

Question 41. Find a particular solution of the differential equation (dx/dy) + xcoty = y^{2}coty + 2y, given that x = 0, when y = π/2Solution:

We have,

(dx/dy) + xcoty = y^{2}coty + 2y………..(i)The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = coty, Q = y^{2}coty + 2y

So,

I.F = e^{∫Pdy}

= e^{∫cotydy}

= e^{log|siny|}

= siny

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(siny) = ∫(y^{2}coty + 2y)sinydy + c

x(siny) = ∫(y^{2}cosy + 2xsiny)dy + c

x(siny) = y^{2}∫cosydx – {(d/dy)y^{2}∫cosydy}dy + ∫2ysinydy + c

x(siny) = y^{2}siny – ∫2ysinydy + ∫2ysinydy + c

x(siny) = y^{2}siny + c

At x = 0, y = π/2

0 = (π/2)^{2}sin(π/2) + c

c = -π^{2}/4

x(siny) = y^{2}siny – π^{2}/4

This is the required solution.

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.