Question

how do I solve x dy/dx+2y= x^-3

Answer 1

in standard form I get
dy/d\[\frac{ dy }{ dx}+ \frac{ 2y }{ x } = x ^{-4}\]

Answer 2

Nevermind I got it :D
the IF is x^2
then \[x^{2} \frac{ dy }{ dx} +\frac{ 2x^2 }{y}=\frac{ 1 }{ x^2}\]
which is then the derivative of \[x^{2}y\]
take the integral of both sides and we get
\[\int\limits \frac{ 1 }{ x^2 }dx =\frac{ -1 }{ x} +c\]
solve for y
\[x^2y= \frac{ -1 }{ x} +c\]
\[y= \frac{ c }{ x^2}-\frac{ 1 }{ x^3 }\]