RD Sharma Solutions for Class 12 Maths Exercise 22.10 Chapter 22 Differential Equations are provided here. Our specialists will teach you many methods to tackle questions in 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.

RD Sharma Solutions for Class 12 Maths Chapter 22 Exercise 10

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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 = e3x … [equation (i)]

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

So, p = 2, Q = e3x

IF = e∫pdx

= e∫2dx

= e2x

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

e2x(dy/dx) + e2x 2y = e2x × e3x

e2x(dy/dx) + e2x 2y = e5x … [because am × an = am + n]

Now, integrating the above equation with respect to x,

ye2x = ∫e5x dx + c

ye2x = (e5x/5) + c

y = (e5x/5) + ce-2x

Question. 2

Solution:

From the question, it is given that,

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

Dividing both sides by 4, we get,

dy/dx + 2y = 5e-3x/4 … [equation (i)]

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

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

IF = e∫pdx

= e∫2dx

= e2x

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

e2x(dy/dx) + e2x 2y = e2x × 5e-3x/4

e2x(dy/dx) + e2x 2y = 5e-x/4

Now, integrating the above equation with respect to x,

ye2x = ∫5e-x/4 dx+ c

ye2x = (-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 = 6ex … [equation (i)]

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

So, p = 2, Q = 6ex

IF = e∫pdx

= e∫2dx

= e2x

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

e2x(dy/dx) + e2x 2y = e2x × 6ex

e2x(dy/dx) + e2x 2y = 6e3x … [because am × an = am + n]

Now, integrating the above equation with respect to x,

ye2x = ∫6e3x dx + c

ye2x = (6/3) e3x + c

y = 2e3x + 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 compared with, (dy/dx) + py = Q

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

IF = e∫pdx

= e∫1dx

= ex

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

ex(dy/dx) + ex y = ex × e-2x

ex(dy/dx) + ex y = e-x … [because am × an = am + n]

Now, integrating the above equation with respect to x,

yex = ∫e-x dx + c

yex = 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 compared with (dy/dx) + py = Q

So, p = – 1/x, Q = 1

IF = e∫pdx

= e∫-1/xdx

= e-log x

= elog(1/x)

= 1/x

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

Y(1/x) = ∫1 (1/x) dx + c

yex = log x + c

y = x log x + cx

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