Question

Solve differential equation (2y-e^x)dx+xdy=0, x>0

First order differential equations
ANSWERED
asked 2020-11-10
Solve differential equation \((2y-e^x)dx+xdy=0\), x>0

Answers (1)

2020-11-11

\((2y-e^x)dx+xdy=0\)
\(2y-e^x+x dy/dx=0\)
\((2y)/x-e^x/x+dy/dx=0\)
\(dy/dx+(2/x)y=e^x/x\)
Above differential equation is form of \(dy/dx+Py=Q\)
Integrating factor will be
\(I.F.= e^{\int Pdx}\)
\(= e^{\int 2/x dx}\)
\(= e^{2 \ln x}\) \(= e^{\ln x^2}\)
\(= x^2\)
Solution will be given as
\(I.F*y= \int I.F*Qdx\)
\(x^2y= \int x^2* e^2/x dx\)
\(x^2y= \int xe^x dx\)
\(x^2y= [x \int e^x dx-\int [d/dx(x) \int e^x dx]] dx\)
\(x^2y= [xe^x-e^x]+c\)
\(y=1/x^2(xe^x-e^x+c)\)

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