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