# Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.4

### 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