Question

Solve differential equationy'+y/x= 3y^(-2) x>0

First order differential equations
ANSWERED
asked 2020-12-13

Solve differential equation \(y'+y/x= 3y^{-2}\) \(x>0\)

Expert Answers (1)

2020-12-14

Dividing both sidesby \(y^{-2}\)
\(y^2y'+ y^3/x=3\)
\(u=y^3\)
Differentiating both sides with respect to x
\(du/dx= 3y^2 dy/dx\)
\(1/3 (du)/dx= y^2 dy/dx\)
Substituting this in the equation \(y^2y'+ y^3/x=3\)
Substituting \(1/3 (du)/dx= y^2 dy/dx\) and \(u=y^3\) in the equation \(y^2y'+ y^3/x=3\)
\(1/3 (du)/dx+ u/x=3\)
\(1/9 (du)/dx+u/(3x)=1\)

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