# Important Questions for Class 12 Chapter 9 - Differential Equation

The Important questions for CBSE class 12th Maths PDF includes complete sets of questions from the topic Sets. It will help you to prepare well for your board exams or any other competitive exams. Further the PDF we provide is Free. Download and Start practicing.

**Important questions based on NCERT syllabus for Class 12 Chapter 9 - Diffrential Equation:**

*Question 1*: Solve the differential equation xlogx dy/dx + y = 2/x logx

*Solution*: Given, the differential equation xlogx dy/dx + y = 2/x logx

Divide xlogx on both side of the equation

dy/dx + y/xlogx = (2/(x^2 logx))logx

dy/dx + y/xlogx = 2/x^2 .......(1)

Equation (1) is of the form

dy/dx + py = Q......(2)

Compare equation 1 and 2

Here P = 1/xlogx and Q = 2/x^2

Integrating factor = IF = e^(∫1/xlogx) = e^(log(logx))

IF = logx [since e^(logx) = x]

Solution of teh above equation is given by

y * IF = ∫(Q * IF) dx + C.......(3)

Substitute IF and Q in the equation 3 we get

ylogx = ∫2/x^2 (log x) dx

We can solve this by using integration by parts

=> ylogx = logx ∫ 2/x^2 dx - ∫ (d/dx (logx). ∫2/x^2 dx)dx

=> ylogx = logx. 2(-1/x)- ∫ 1/x. 2(-1/x) dx(1)

=> ylogx = -2/x logx - ∫2/x(-1/x) dx

=> ylogx = -2/x logx + ∫2/x^2 dx

=> ylogx = -2/x logx - 2/x + C

Which is the required solution.

*Question 2*: Find the general solution of the equation (x^2 -1)dy/dx + 2xy = 2/x^2 - 1

*Solution*: Given the differential equation (x^2 -1)dy/dx + 2xy = 2/x^2 - 1

Divide (x^2 - 1) on both sides of the equation

dy/dx + 2xy/x^2 - 1 = 2/(x^2 -1)(x^2-1)

dy/dx + (2x/x^2 - 1)y = 2/(x^2 -1)^2........(1)

Which is linear differential equation

Compare equation (1) with dy/dx + py = Q

Here P = 2x/x^2 - 1 and Q = 2/(x^2-1)^2

Integrating factor = IF = e^{∫(2x/x^2-1)} = e^{log|x^2-1|} = x^2 - 1

Hence general solution becomes

y IF = ∫Q * IF dx + C

=> y(x^2-1) = ∫2/(x^2-1)^2 * (x^2-1) dx + C

=> y(x^2-1) = ∫2/(x^2-1) dx + C

=> y(x^2-1) = log|(x-1)/(x+1)| + C is the required general solution.