Question

Find the integrating factor for the following differential equation:`x logx dy/dx+y=2log x`

Answer 1

Consider the given differential equation:

`x logx dy/dx+y=2log x`

Dividing the above equation by xlogx, we have,

`(x logx)/(x logx)dy/dx+y/(x logx)=(2log x)/(x logx)`

`=>dy/dx+y/(x logx)=1/x ........(1)`

Consider the general linear differential equation

`dy/dx+Py=Q,` where P and Q are functions of x.

Comparing equation (1) and the general equation, we have,

`P(x)=1/xlogx and Q(x)=2/x`

The integrating factor is given by the formula `e^(intPdx)`

Thus `I.F=e^(intPdx)=e^(intdx/(xlogx))`

Consider `I=int dx/(xlogx)`

Substituting logx=t; dx/x=dt

Thus `I=intdt/t=log(t)=log(logx)`

Hence ` I.F=e^(intdx/(xlogx))=e^(log(logx))=logx`