Question

xdy+ydx=(x^3y^6)dx

answer: 1/y^5=-5x^3/2+(c1(x^5))

Answer 1

1.divide by dx
2.divide by xy^6
3. put \[y ^{-5}=t,-5y ^{-6}\frac{ dy }{ dx }=\frac{ dt }{ dx }\]

Answer 2

Another approach (though essentially the same as @Princer_Jones's ):
\[\begin{align*}
xy'+y&=x^3y^6&\text{set }t=yx\\\\
t'&=t^3y^3\\\\
t'&=\frac{t^6}{x^3}
\end{align*}\]

Answer 3

i am trying to understand

Answer 4

@rubcksy there will not be any negative in the rhs..

Answer 5

\[y ^{-6}\frac{ dy }{ dx }+\frac{ 1 }{ x }y ^{-5}=x^2\]
put \[y ^{-5}=t,-5y ^{-6}\frac{ dy }{ dx }=\frac{ dt }{ dx }\]
\[y ^{-6}\frac{ dy }{ dx }=-\frac{ 1 }{ 5 }\frac{ dt }{ dx }\]
\[-\frac{ 1 }{ 5 }\frac{ dt }{ dx }+\frac{ 1 }{ x }t=x^2\]
\[\frac{ dt }{ dx }+\frac{ -5 }{ x }t=-5x^2\]
\[I.F=e ^{\int\limits \frac{ -5 }{ x }}dx=e ^{-5\ln x}=e ^{\ln x ^{-5}}=x ^{-5}\]
C.S. is \[tx ^{-5}=\int\limits x^2x ^{-5}dx=\int\limits x ^{-3}dx+c\]
complete it.