Question

Solve differential equationdy/dx= x/(ye^(x+y^2))

First order differential equations
ANSWERED
asked 2020-12-28

Solve differential equation \(dy/dx= x/(ye^{x+y^2})\)

Answers (1)

2020-12-29

\(dy/dx= x/(ye^{x+y^2})\)
\(=> dy/dx= x/(ye^x*e^{y^2})\)
\(=> ye^(y^2)dy= x/e^x dx\)
\(ye^{y^2}dy= xe^{-x} dx\)
\(\int e^{y^2}ydy=\int xe^{-x} dx+c\)
where C is integrating constant
\(\int e^{y^2}ydy=\int xe^{-x} dx+c\)
\(y^2=t\)
\(2ydy=dt\)
\(y dy= (dt)/2\)
\(\Rightarrow\frac{1}{2}\int e^t dt=\int xe^{-x} dx+c\)(integrating by part)
\(\Rightarrow\frac{1}{2}e^t=-xe^{-x}+\int e^{-x} dx+c\)
\(1/2 e^{y^2}= -xe^{-x}+(-e^{-x} +c \therefore (t=y^2))\)
\(1/2 y^2= -e^-x [x+1]+c\)
\(1/2 y^2+e^-n(x+1)=c\)
\(1/2 y^2= -e^-x(x+1)+c\)
\(y^2/2+e^-x(x+1)=C\)

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