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 = e⁰ = 1 

So, y(0) = 1

Question 3. For each of the following initial value problems verify that the accompanying function is a solution: (d²y/dx²) + 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,

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

Question 4. For each of the following initial value problems verify that the accompanying function is a solution: d²y/dx² – (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,

d²y/dx² = e^x

d²y/dx² – e^x = 0        

d²y/dx² – (dy/dx) = 0

Thus, y = e^x + 1 satisfy the given differential equation.

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

y'(0) = e⁰ = 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: (d²y/dx²) + 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,

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

Question 7. For each of the following initial value problems verify that the accompanying function is a solution: (d²y/dx²) – 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,

d²y/dx² = e^x + e^-x

d²y/dx² = y

d²y/dx² – y = 0

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

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

y'(0) = e⁰ – e^-0 = 0

Question 8. For each of the following initial value problems verify that the accompanying function is a solution: (d²y/dx²) – 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,

d²y/dx² = e^x + 4e^2x

d²y/dx² = 3(e^x + 2e^2x) – 2(e^x + e^2x)

(d²y/dx²) = 3(dy/dx) – 2y

(d²y/dx²) – 3(dy/dx) + 2y = 0

Thus, y = e^x + e^2x satisfy the given differential equation.

If x = 0, y(0) = e⁰ + e⁰ = 1 + 1 = 2

y'(0) = e⁰ + 2e⁰ = 1 + 2 = 3

Question 9. For each of the following initial value problems verify that the accompanying function is a solution: (d²y/dx²) – 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,

d²y/dx² = xe^x + e^x + 2e^x 

d²y/dx² = xe^x + e^x + 2e^x + xe^x + e^x – xe^x – e^x

d²y/dx² = 2(xe^x + e^x) – (xe^x + e^x)

(d²y/dx²) = 2(dy/dx) – y

(d²y/dx²) – 2(dy/dx) + y = 0

Thus, y = xe^x + e^x satisfy the given differential equation.

If x = 0, y(0) = 0e⁰ + e⁰ = 1

y'(0) = 0e⁰ + 2e⁰ = 2

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