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Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 22 |

Exercise | 22.2 |

Category | RD Sharma Solutions |

**RD Sharma Class 12 Ex 22.2 Solutions Chapter 22 Differential Equations**

### Question 1. Form the differential equation of the family of curves represented by y^{2 }= (x – c)^{3}

**Solution:**

y

^{2 }= (x – c)^{3}On differentiating the given equation w.r.t x,

2y(dy/dx) = 3(x – c)

^{2}(x – c)

^{2 }= (2y/3)(dy/dx)(x – c) = [(2y/3)(dy/dx)]

^{1/2 }-(1)On putting the value of (x – c) in equation (1), we get

y

^{2}= [(2y/3)(dy/dx)]^{3/2}On squaring both side, we get

y

^{4 }= [(2y/3)(dy/dx)]^{3}y

^{4}= (8y^{3}/27)(dy/dx)^{3}27y

^{4 }= 8y^{3}(dy/dx)^{3}27y = 8(dy/dx)

^{3}

### Question 2. Form the differential equation corresponding to y = e^{mx} by eliminating m.

**Solution:**

y = e

^{mx }-(1)On differentiating the given equation w.r.t x,

dy/dx = me

^{mx }-(2)From eq(1), we get

y = e

^{mx}logy = mx

m = (logy/x)

Now, put the value of m in equation(2), we get

x(dy/dx) = ylogy

### Question 3. Form the differential equations from the following primitives where constants are arbitrary.

### (i) y^{2 }= 4ax

**Solution:**

y

^{2 }= 4ax^{ }-(1)y

^{2}/4x = aOn differentiating the given equation w.r.t x,

2y(dy/dx) = 4a

^{ }-(2)Now, put the value of a in equation(2), we get

2y(dy/dx) = 4(y

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

^{2}/x2x(dy/dx) = y

### (ii) y = cx + 2c^{2 }+ c^{3}

**Solution:**

y = cx + 2c

^{2 }+ c^{3 }-(1)On differentiating the given equation w.r.t x,

dy/dx = c

^{ }-(2)Now, put the value of c in equation(1), we get

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

^{2 }+ (dy/dx)^{3}

### (iii) xy = a^{2 }

**Solution:**

xy = a

^{2 }-(1)On differentiating the given equation w.r.t x,

x(dy/dx) + y = 0

### (iv) y = ax^{2 }+ bx + c

**Solution:**

y = ax

^{2 }+ bx + c^{ }-(1)On differentiating the given equation w.r.t x,

dy/dx = 2ax + b

^{ }-(2)Again differentiating the given equation w.r.t x,

d

^{2}y/dx^{2 }= 2a^{ }-(3)Again, differentiating the given equation w.r.t x, we get

d

^{3}y/dx^{3 }= 0

### Question 4. Find the differential equation of the family of curves y = Ae^{2x }+ Be^{-2x} where A and B are arbitrary constants.

**Solution:**

y = Ae

^{2x }+ Be^{-2x }-(1)On differentiating the given equation w.r.t x,

(dy/dx) = 2Ae

^{2x }– 2Be^{-2x }-(2)Again, differentiating the given equation w.r.t x,

d

^{2}y/dx^{2 }= 4Ae^{2x }+ 4Be^{-2x}d

^{2}y/dx^{2 }= 4(Ae^{2x }+ Be^{-2x})d

^{2}y/dx^{2 }= 4y

### Question 5. Find the differential equation of the family of curves, x = Aconst + Bsinnt where A and B are arbitrary constants.

**Solution:**

x = Acosnt + Bsinnt

^{ }-(1)On differentiating the given equation w.r.t x,

dy/dx = -Ansinnt + Bncosnt

^{ }-(2)Again, differentiating the given equation w.r.t x,

d

^{2}y/dx^{2 }= -An^{2}cosnt – Bn^{2}sinntd

^{2}y/dx^{2 }= -n^{2}(Acosnt + Bsinnt)d

^{2}y/dx^{2 }+ n^{2}x = 0

### Question 6. Form the differential equation corresponding to y^{2} = a(b – x^{2}) by eliminating a and b.

**Solution:**

y^{2 }= a(b – x^{2})

On differentiating the given equation w.r.t x,

2y(dy/dx) = a(0 – 2x)

Again, differentiating the given equation w.r.t x,

x[

### Question 7. Form the differential equation corresponding to y^{2 }– 2ay + x^{2 }= a^{2 }by eliminating a.

**Solution:**

y^{2 }– 2ay + x^{2 }= a^{2 } -(1)

On differentiating the given equation w.r.t x,

2y(dy/dx) – 2a(dy/dx) + 2x = 0

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

^{ } -(2)

Let us considered dy/dx = y’

On putting the value of ‘a’ in the eq(1), we get

On solving this equation, we get

(x^{2 }– 2y^{2})y’^{2 }– 4xyy’ – x^{2 }= 0

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

### Question 8. Form the differential equation corresponding to (x – a)^{2 }+ (y – b)^{2 }= r^{2 }by eliminating a and b.

**Solution:**

(x – a)^{2 }+ (y – b)^{2 }= r^{2 } -(1)

On differentiating the given equation w.r.t x,

2(x – a) + 2(y – b)(dy/dx) = 0

(x – a) + (y – b)(dy/dx) = 0^{ } -(2)

Again, differentiating the given equation w.r.t x,

1 + (y – b)(d^{2}y/dx^{2}) + (dy/dx)(dy/dx) = 0

^{ } -(3)

On putting the value of (y – b) in eq(2),

(x – a)(d^{2}y/dx^{2}) – (dy/dx)^{3 }– (dy/dx) = 0

^{ } -(4)

On putting the value of (x – a) and (y – b) in eq(1)

Put (dy/dx) = y’ and d^{2}y/dx^{2} = y”

y’^{2}(1 + y’^{2})^{2 }+ (1 + y’^{2})^{2 }= r^{2}y”^{2}

### Question 9. Find the differential equation of all the circles which pass through the origin and whose centres lie on-axis.

**Solution:**

Equation of the circle is (x – a)^{2 }+ (y – b)^{2 }= r^{2 }

Here, a and b are the centre of the circle.

(x – a)^{2 }+ (y – b)^{2 }= r^{2 } -(1)

When the centre lies on the y-axis, so a = 0

x^{2 }+ (y – b)^{2 }= r^{2 } -(2)

So, when the circle is passing through origin, so the equation is

0^{2 }+ b^{2 }= r^{2 } -(3)

x^{2 }+ (y – b)^{2 }= r^{2}

On squaring both side, we have

x^{2 }+ y^{2 }– 2yr + r^{2 }= r^{2} -(Since r^{2 }= b^{2})

2yr = x^{2 }+ y^{2}

r = (x^{2 }+ y^{2})(2y)

On differentiating the equation(1) w.r.t x, we get

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

0 = y^{2}(dy/dx) – x^{2}(dy/dx) + 2xy

(x^{2 }– y^{2})(dy/dx) = 2xy

### Question 10. Find the differential equation of all the circles which pass through the origin and whose centres lie on the x-axis.

**Solution:**

Equation of the circle is (x – a)^{2 }+ (y – b)^{2 }= r^{2}

Here, a and b are the centre of the circle.

When the centre lies on the x-axis, so b = 0

(x – a)^{2 }+ y^{2 }= r^{2 } -(1)

When the circle is passing through origin, so the equation is

a^{2 }+ 0^{2 }= r^{2}^{ } -(2)

(x – a)^{2 }+ y^{2 }= r^{2}

On squaring both side, we get,

x^{2 }– 2ax + a^{2 }+ y^{2 }= r^{2}

x^{2 }+ y^{2 }– 2xr + r^{2 }= r^{2} -(Since r^{2 }= a^{2})

2xr = x^{2 }+ y^{2 }

r = (x^{2 }+ y^{2})(2x)^{ } -(3)

On differentiating the equation w.r.t x, we get

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

(x^{2 }– y^{2}) + 2xy(dy/dx) = 0

### Question 11. Assume that a raindrop evaporates at a rate proportional to its surface area. Form a differential equation involving the rate of change of the radius of the raindrop.

**Solution:**

Let us considered ‘r’ be the radius of rain drop, volume of the drop be ‘V’ and area of the drop be ‘A’

(dV/dt) proportional to A

(dV/dt) – kA -(V decreases with increasing in t so negative sing)

Here, k is proportionality constant,

= -k(4π r

^{2})4πr

^{2}(dr/dt) = -k(4πr^{2})(dr/dt) = -k

### Question 12. Find the differential equation of all the parabolas with latus rectum 4a’ and whose axes are parallel to the x-axis.

**Solution:**

Equation of parabola whose area is parallel to x-axis and vertices at (h, k).

(y – k)^{2 }= 4a(x – h) -(1)

On differentiating w.r.t x,

2(y – k)(dy/dx) = 4a

(y – k)(dy/dx) = 2a

-(2)

Again, differentiating w.r.t x,

d^{2}y/dx^{2}(y – k) + (dy/dx)(dy/dx) = 0

2a(d^{2}y/dx^{2}) + (dy/dx)^{3 }= 0

### Question 13. Show that the differential equation of which is a solution, is (dy/dx) + 2xy = 4x^{3}

**Solution:**

-(1)

On differentiating w.r.t x,

On adding 2xy in R.H.S and L.H.S,

On putting the value of y in above equation,

=

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

### Question 14. From the differential equation having y = (sin^{-1}x)^{2} + A cos^{-1}x + B, where A and B are arbitrary constants, as its general solution.

**Solution:**

y = (sin^{-1}x)^{2} + A cos^{-1}x + B

On differentiating w.r.t x,

Again, on differentiating w.r.t x,

### Question 15. Form the differential equation of the family of curves represented by the equation (a being the parameter)

### (i) (2x + a)^{2 }+ y^{2 }= a^{2}

**Solution:**

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

On differentiating w.r.t x,

2(2x + a) + 2y(dy/dx) = 0

(2x + a) + y(dy/dx) = 0

a = -2x – y(dy/dx) -(2)

On putting the value of ‘a’ in eq(1), we have

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

y^{2 }– 4x^{2 }– 4xy(dy/dx) = 0

### (ii) (2x – a)^{2 }– y^{2 }= a^{2}

**Solution:**

(2x – a)^{2 }– y^{2 }= a^{2}

4x^{2 }– 4ax + a^{2 }– y^{2 }= a^{2 }

4ax = 4x^{2 }– y^{2}

a = (4x^{2 }– y^{2})/4x

On differentiating w.r.t x,

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

### (iii) (x – a)^{2 }+ 2y^{2 }= a^{2 }

**Solution:**

(x – a)^{2 }+ 2y^{2 }= a^{2} -(1)

On differentiating w.r.t x,

2(x – a) + 4y(dy/dx) = 0

(x – a) + 2y(dy/dx) = 0

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

On putting the value of a in eq(1)

2y^{2 }– 4xy(dy/dx) – x^{2 }= 0

### Question 16. Represent the following families of curves by forming the corresponding differential equations (a, b being parameters):

### (i) x^{2 }+ y^{2 }= a^{2}

**Solution:**

x

^{2 }+ y^{2 }= a^{2}On differentiating w.r.t x,

2x + 2y(dy/dx) = 0

x + y(dy/dx) = 0

### (ii) x^{2 }– y^{2 }= a^{2}

**Solution:**

x

^{2 }– y^{2 }= a^{2}On differentiating w.r.t x,

2x – 2y(dy/dx) = 0

x – y(dy/dx) = 0

### (iii) y^{2 }= 4ax

**Solution:**

y^{2 }= 4ax

(y2/x) = 4a

On differentiating w.r.t x,

2xy(dy/dx) – y^{2 }= 0

2x(dy/dx) – y = 0

### (iv) x^{2 }+ (y – b)^{2 }= 1

**Solution:**

x^{2 }+ (y – b)^{2 }= 1 -(1)

On differentiating w.r.t x,

2x + 2(y – b)(dy/dx) = 0

On putting the value of (y – b) in eq(1)

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

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

### (v) (x – a)^{2 }– y^{2 }= 1

**Solution:**

(x – a)

^{2 }– y^{2 }= 1 -(1)On differentiating w.r.t x,

2(x – a) – 2y(dy/dx) = 0

(x – a) – y(dy/dx) = 0

(x – a) = y(dy/dx)

On putting the value of (y – b) in eq(i), we get

y

^{2}(dy/dx)^{2 }– y^{2 }= 1y

^{2}[(dy/dx)2 – 1] = 1

### (vi)

**Solution:**

We have,

-(1)

{(bx)^{2 }– (ay)^{2}} = (ab)^{2} -(2)

On differentiating w.r.t x,

2xb^{2 }– 2a^{2}y(dy/dx) = 0

xb^{2 }– a^{2}y(dy/dx) = 0 -(3)

Again, differentiating w.r.t x,

On putting the value of b^{2} in equation(3), we get

xb^{2 }– a^{2}y(dy/dx) = 0

### (vii) y^{2 }= 4a(x – b)

**Solution:**

We have,

y^{2 }= 4a(x – b)

On differentiating w.r.t x,

2y(dy/dx) = 4a

Again differentiating w.r.t x,

[(dy/dx)^{2 }+ y(d^{2}y/dx^{2})] = 0

### (viii) y = ax^{3}

**Solution:**

We have,

y = ax

^{3}-(1)On differentiating w.r.t x,

(dy/dx) = 3ax

^{2}From eq(1),

a=(y/x

^{3}-(1)On putting the value of a in eq(1)

dy/dx = 3(y/x

^{3}) × x^{2}x(dy/dx) = 3y

### (ix) x^{2 }+ y^{2 }= ax^{3}

**Solution:**

We have,

x^{2 }+ y^{2 }= ax^{3}

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

On differentiating w.r.t x,

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

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

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

### (x) y = e^{ax}

**Solution:**

We have,

y = e

^{ax}-(1)On differentiating w.r.t x,

dy/dx = ae

^{ax}dy/dx = ay -(2)

y = e

^{ax }On taking log both side, we get

logy = ax

a = (logy/x)

Now, put the value of ‘a’ in eq(2)

(dy/dx) = logy/x) × y

x(dy/dx) = ylogy

### Question 17. Form the differential equation representing the family of ellipses having foci on the x-axis and the centre at the origin.

**Solution:**

We have,

Equation of ellipse having foci on the x-axis,

-(where a > b)

(bx)^{2 }+ (ay)^{2 }= (ab)^{2} -(1)

On differentiating above equation w.r.t x,

2b^{2}x + 2a^{2}y(dy/dx) = 0

b^{2}x + a^{2}y(dy/dx) = 0 -(2)

Again, differentiating w.r.t x,

On putting the value of b^{2} in eq(2),

xb^{2 }+ a^{2}y(dy/dx) = 0

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

### Question 18. Form the differential equation of the family of hyperbolas having foci on the X-axis and centre at the origin

**Solution:**

We have,

Equation of a hyperbola having a Centre at the origin and foci along x-axis

-(1)

(bx)^{2}-(ay)^{2}=(ab)^{2} -(2)

On differentiating above equation w.r.t x,

2xb^{2}-2a^{2}y(dy/dx)=0

xb^{2}-a^{2}y(dy/dx)=0 -(3)

Again, differentiating above equation w.r.t x,

Putting the value of b^{2} in equation(3),

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

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

This is required differential equation.

### Question 19. Form the differential equation of the family of circles in the second quadrant and touching the coordinate axis.

**Solution:**

We have,

Let (-a, a) be the coordinates of the centre of circle

So, the equation of circle is given by,

(x + a)^{2 }+ (y – b)^{2 }= a^{2} -(1)

x^{2 }+ 2ax + a^{2 }+ y^{2 }– 2ay + a^{2 }= 0 -(2)

On differentiating above equation w.r.t x,

2x + 2a + 2y(dy/dx) – 2a(dy/dx) = 0

x + a + y(dy/dx) – a(dy/dx) = 0

On substituting the value of ‘a’ in eq(2)

Let, (dy/dx) = p

[xp – x + x + yp]^{2 }+ [yp – y – x – yp]^{2 }= [x + yp]^{2}

(x + y)^{2}p^{2 }+ (x + y)^{2 }= (x + yp)^{2}

(x + y)^{2}[p^{2 }+ 1] = (x + yp)^{2} -(where (dy/dx) = p)

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