Important Questions for Class 12 Chapter 9 - Differential Equation

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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.