RD Sharma Solutions Class 12 Differential Equations Exercise 22.10

RD Sharma Solutions for Class 12 Maths Exercise 22.10 Chapter 22 Differential Equations are provided here. Our specialists will guide you in many methods to tackle questions on the exam and manage time in the examination. Students can refer to RD Sharma Solutions for Class 12 Maths Chapter 22 and practise various questions daily to perform well in the final exams.

Download PDF of RD Sharma Solutions for Class 12 Maths Chapter 22 Exercise 10

Access RD Sharma Solutions for Class 12 Maths Chapter 22 Exercise 10

EXERCISE 22.10

Question. 1

Solution:

From the question it is given that,

(dy/dx) + 2y = e^3x … [equation (i)]

The given linear differential equation is comparing with, (dy/dx) + py = Q

So, p = 2, Q = e^3x

IF = e^∫pdx

= e^∫2dx

= e^2x

Then, multiplying both side of equation (i) by IF,

e^2x(dy/dx) + e^2x 2y = e^2x × e^3x

e^2x(dy/dx) + e^2x 2y = e^5x … [because a^m × aⁿ = a^{m + n}]

Now, integrating the above equation with respect to x,

ye^2x = ∫e^5x dx + c

ye^2x = (e^5x/5) + c

y = (e^5x/5) + ce^-2x

Question. 2

Solution:

From the question it is given that,

4(dy/dx) + 8y = 5e^-3x

Dividing both side by 4 we get,

dy/dx + 2y = 5e^-3x/4 … [equation (i)]

The given linear differential equation is comparing with, (dy/dx) + py = Q

So, p = 2, Q = 5e^-3x/4

IF = e^∫pdx

= e^∫2dx

= e^2x

Then, multiplying both side of equation (i) by IF,

e^2x(dy/dx) + e^2x 2y = e^2x × 5e^-3x/4

e^2x(dy/dx) + e^2x 2y = 5e^-x/4

Now, integrating the above equation with respect to x,

ye^2x = ∫5e^-x/4 dx+ c

ye^2x = (-5/4) e^-x + c

y = (-5/4) e^-3x + ce^-2x

Question. 3

Solution:

From the question it is given that,

(dy/dx) + 2y = 6e^x … [equation (i)]

The given linear differential equation is comparing with, (dy/dx) + py = Q

So, p = 2, Q = 6e^x

IF = e^∫pdx

= e^∫2dx

= e^2x

Then, multiplying both side of equation (i) by IF,

e^2x(dy/dx) + e^2x 2y = e^2x × 6e^x

e^2x(dy/dx) + e^2x 2y = 6e^3x … [because a^m × aⁿ = a^{m + n}]

Now, integrating the above equation with respect to x,

ye^2x = ∫6e^3x dx + c

ye^2x = (6/3) e^3x + c

y = 2e^3x + ce^-2x

Question. 4

Solution:

From the question it is given that,

(dy/dx) + y = e^-2x … [equation (i)]

The given linear differential equation is comparing with, (dy/dx) + py = Q

So, p = 1, Q = e^-2x

IF = e^∫pdx

= e^∫1dx

= e^x

Then, multiplying both side of equation (i) by IF,

e^x(dy/dx) + e^x y = e^x × e^-2x

e^x(dy/dx) + e^x y = e^-x … [because a^m × aⁿ = a^{m + n}]

Now, integrating the above equation with respect to x,

ye^x = ∫e^-x dx + c

ye^x = e^-x/-1 + c

y = -e^-2x + ce^-x

Question. 5

Solution:

From the question it is given that,

x(dy/dx) = x + y

dy/dx = (x + y)/x

dy/dx = 1 + (y/x) … [equation (i)]

The given linear differential equation is comparing with, (dy/dx) + py = Q

So, p = – 1/x, Q = 1

IF = e^∫pdx

= e^∫-1/xdx

= e^{-log x}

= e^log(1/x)

= 1/x

Then, multiplying both side of equation (i) by IF,

Y(1/x) = ∫1 (1/x) dx + c

ye^x = log x + c

y = x log x + cx