Here we provide RD Sharma Class 12 Ex 22.4 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 22.4 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 | 22.4 |

Category | RD Sharma Solutions |

**RD Sharma Class 12 Ex 22.4 Solutions Chapter 22 Differential Equations**

### Question 1. For each of the following initial value problems verify that the accompanying function is a solution: x(dy/dx) = 1, y(1) = 0

### Function: y = logx

**Solution:**

We have,

y = logx -(1)

On differentiating eq(1) w.r.t x,

dy/dx = (1/x)

x(dy/dx) = 1

Thus, y = logx satisfy the given differential equation.

If x = 1, y = log(1) = 0

So, y(1) = 0

### Question 2. For each of the following initial value problems verify that the accompanying function is a solution: (dy/dx) = y, y(0) = 0

### Function: y = e^{x }

**Solution:**

We have,

y = e

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

dy/dx = e

^{x}(dy/dx) = y

Thus, y = e

^{x}satisfy the given differential equation.If x = 0, y = e

^{0 }= 1So, y(0) = 1

### Question 3. For each of the following initial value problems verify that the accompanying function is a solution: (d^{2}y/dx^{2}) + y = 0, y(0) = 0, y'(0) = 1

### Function: y = sinx

**Solution:**

We have,

y = sinx -(1)

On differentiating eq(1) w.r.t x,

(dy/dx) = cosx -(2)

Again, differentiating eq(2) w.r.t x,

d

^{2}y/dx^{2 }= -sinxd

^{2}y/dx^{2 }+ sinx = 0Thus, y = sinx satisfy the given differential equation.

If x = 0, y(0) = sin(0) = 0

y'(0) = cos(0) = 1

### Question 4. For each of the following initial value problems verify that the accompanying function is a solution: d^{2}y/dx^{2 }– (dy/dx) = 0, y(0) = 2, y'(0) = 1

### Function: y = e^{x }+ 1

**Solution:**

We have,

y = e

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

(dy/dx) = e

^{x}-(2)Again differentiating eq(2) w.r.t x,

d

^{2}y/dx^{2 }= e^{x}d

^{2}y/dx^{2 }– e^{x }= 0d

^{2}y/dx^{2 }– (dy/dx) = 0Thus, y = e

^{x }+ 1 satisfy the given differential equation.If x = 0, y(0) = e

^{0 }+ 1, y(0) = 1 + 1 = 2y'(0) = e

^{0 }= 1

### Question 5. For each of the following initial value problems verify that the accompanying function is a solution: (dy/dx) + y = 2

### Function: y = e^{-x }+ 2

**Solution:**

We have,

y = e

^{-x }+ 2 -(1)On differentiating eq(i) w.r.t x,

(dy/dx) = -e

^{-x}(dy/dx) + e

^{-x }= 0(dy/dx) + (y – 2) = 0

(dy/dx) + y = 2

Thus, y = e

^{-x }+ 2 satisfy the given differential equation.If x = 0, y(0) = e

^{-0 }+ 2 = 1 + 2 = 3

### Question 6. For each of the following initial value problems verify that the accompanying function is a solution: (d^{2}y/dx^{2}) + y = 0, y(0) = 1, y'(0) = 1

### Function: y = sinx + cosx

**Solution:**

We have,

y = sinx + cosx -(1)

On differentiating eq(i) w.r.t x,

dy/dx = cosx – sinx -(2)

Again differentiating eq(ii) w.r.t x,

d

^{2}y/dx^{2 }= -sinx – cosxd

^{2}y/dx^{2 }= -(sinx + cosx)(d

^{2}y/dx^{2}) + y = 0Thus, y = sinx + cosx satisfy the given differential equation.

If x = 0, y(0) = sin0 + cos0 = 1

y'(0) = cos0 – sin0 = 1

### Question 7. For each of the following initial value problems verify that the accompanying function is a solution: (d^{2}y/dx^{2}) – y = 0, y(0) = 2, y'(0) = 0

### Function: y = e^{x }+ e^{-x}

**Solution:**

We have,

y = e

^{x }+ e^{-x}-(1)On differentiating eq(i) w.r.t x,

dy/dx = e

^{x }– e^{-x}-(2)Again differentiating eq(2) w.r.t x,

d

^{2}y/dx^{2 }= e^{x }+ e^{-x}d

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

^{2}y/dx^{2 }– y = 0Thus, y = e

^{x }+ e^{-x}satisfy the given differential equation.If x = 0, y(0) = e

^{0 }+ e^{-0 }= 1 + 1 = 2y'(0) = e

^{0 }– e^{-0 }= 0

### Question 8. For each of the following initial value problems verify that the accompanying function is a solution: (d^{2}y/dx^{2}) – 3(dy/dx) + 2y = 0, y(0) = 2, y'(0) = 3

### Function: y = e^{x }+ e^{2x}

**Solution:**

We have,

y = e

^{x }+ e^{2x}-(1)On differentiating eq(1) w.r.t x,

dy/dx = e

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

d

^{2}y/dx^{2 }= e^{x }+ 4e^{2x}d

^{2}y/dx^{2 }= 3(e^{x }+ 2e^{2x}) – 2(e^{x}+ e^{2x})(d

^{2}y/dx^{2}) = 3(dy/dx) – 2y(d

^{2}y/dx^{2}) – 3(dy/dx) + 2y = 0Thus, y = e

^{x }+ e^{2x}satisfy the given differential equation.If x = 0, y(0) = e

^{0 }+ e^{0 }= 1 + 1 = 2y'(0) = e

^{0 }+ 2e^{0 }= 1 + 2 = 3

### Question 9. For each of the following initial value problems verify that the accompanying function is a solution: (d^{2}y/dx^{2}) – 2(dy/dx) + y = 0, y(0) = 1, y'(0) = 2

### Function: y = xe^{x }+ e^{x }

**Solution:**

We have,

y = xe

^{x }+ e^{x}-(1)On differentiating eq(1) w.r.t x,

dy/dx = xe

^{x }+ e^{x }+ e^{x }dy/dx = xe

^{x }+ 2e^{x}-(2)Again differentiating eq(2) w.r.t x,

d

^{2}y/dx^{2 }= xe^{x }+ e^{x }+ 2e^{x }d

^{2}y/dx^{2 }= xe^{x }+ e^{x }+ 2e^{x }+ xe^{x }+ e^{x }– xe^{x }– e^{x}d

^{2}y/dx^{2 }= 2(xe^{x }+ e^{x}) – (xe^{x }+ e^{x})(d

^{2}y/dx^{2}) = 2(dy/dx) – y(d

^{2}y/dx^{2}) – 2(dy/dx) + y = 0Thus, y = xe

^{x }+ e^{x}satisfy the given differential equation.If x = 0, y(0) = 0e

^{0 }+ e^{0 }= 1y'(0) = 0e

^{0 }+ 2e^{0 }= 2

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