Question

Solve the differential equationx dy/dx-y= x^2lnx

First order differential equations
ANSWERED
asked 2020-11-16

Solve the differential equation \(x dy/dx-y= x^2\ln x\)

Answers (1)

2020-11-17

Divide both sides by x and further simplify it
\(x/x dy/dx-y/x= (x^2 \ln x)/x\)
\(dy/dx-y/x= x\ln x\)
It is of the form \(dy/dx+P(x)y= Q(x)\)
Now find the integrating factor by using \((I.F.)= e^{\int P(x)dx}\)
Hence, \(I.F.= e^{-\int 1/x dx}\)
\(= e^{-\ln x}\)
\(= e^{\ln 1/x}\)
\(= 1/x\)
Hence, \(I.F.= 1/x\)
\(y*(I.F.)= \int (I.F.)*Q(x)dx+c\)
Hence, \(y*1/x= \int 1/x * x\ln x dx+c\)
\(y*1/x= \int \ln x dx+c\)
\(y*1/x= \int \ln x*1dx+c\)
\(y*1/x= \ln x+x+c\)
Simplify
\(y*1/x=x(\ln x+1)+c\)
\(y=x^2(\ln x+1)+cx\)

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