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

### 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

### Function: y = ex

Solution:

We have,

y = ex          -(1)

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

dy/dx = ex

(dy/dx) = y

Thus, y = ex satisfy the given differential equation.

If x = 0, y = e0 = 1

So, 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,

d2y/dx2 = -sinx

d2y/dx2 + sinx = 0

Thus, y = sinx satisfy the given differential equation.

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

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

### Function: y = ex + 1

Solution:

We have,

y = ex + 1           -(1)

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

(dy/dx) = ex           -(2)

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

d2y/dx2 = ex

d2y/dx2 – ex = 0

d2y/dx2 – (dy/dx) = 0

Thus, y = ex + 1 satisfy the given differential equation.

If x = 0, y(0) = e0 + 1, y(0) = 1 + 1 = 2

y'(0) = e0 = 1

### 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

### 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,

d2y/dx2 = -sinx – cosx

d2y/dx2 = -(sinx + cosx)

(d2y/dx2) + y = 0

Thus, y = sinx + cosx satisfy the given differential equation.

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

y'(0) = cos0 – sin0 = 1

### Function: y = ex + e-x

Solution:

We have,

y = ex + e-x           -(1)

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

dy/dx = ex – e-x           -(2)

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

d2y/dx2 = ex + e-x

d2y/dx2 = y

d2y/dx2 – y = 0

Thus, y = ex + e-x satisfy the given differential equation.

If x = 0, y(0) = e0 + e-0 = 1 + 1 = 2

y'(0) = e0 – e-0 = 0

### Function: y = ex + e2x

Solution:

We have,

y = ex + e2x           -(1)

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

dy/dx = ex + 2e2x           -(2)

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

d2y/dx2 = ex + 4e2x

d2y/dx2 = 3(ex + 2e2x) – 2(ex + e2x)

(d2y/dx2) = 3(dy/dx) – 2y

(d2y/dx2) – 3(dy/dx) + 2y = 0

Thus, y = ex + e2x satisfy the given differential equation.

If x = 0, y(0) = e0 + e0 = 1 + 1 = 2

y'(0) = e0 + 2e0 = 1 + 2 = 3

### Function: y = xex + ex

Solution:

We have,

y = xex + ex            -(1)

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

dy/dx = xex + ex + e

dy/dx = xex + 2ex           -(2)

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

d2y/dx2 = xex + ex + 2e

d2y/dx2 = xex + ex + 2ex + xex + ex – xex – ex

d2y/dx2 = 2(xex + ex) – (xex + ex)

(d2y/dx2) = 2(dy/dx) – y

(d2y/dx2) – 2(dy/dx) + y = 0

Thus, y = xex + ex satisfy the given differential equation.

If x = 0, y(0) = 0e0 + e0 = 1

y'(0) = 0e0 + 2e0 = 2

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