Question

[DiffEq] Find the general solution to the first order linear differential equation.

\[(x^4+2y) dx - x dy = 0\]

Answer 1

\[x^4+2y=x \frac{ dy }{ dx }\]
\[\frac{ dy }{ dx }-\frac{ 2 }{ x }y=x^3\]
\[I.F.=e ^{\int\limits \frac{ -2 }{ x }dx}=e ^{-2\ln x}=e ^{\ln x ^{-2}}=x ^{-2}=\frac{ 1 }{ x^2 }\]
\[C.S.~is~y \times \frac{ 1 }{ x^2 }=\int\limits x^4 \times \frac{ 1 }{ x^2 }dx+c\]
\[\frac{ y }{ x^2 }=\int\limits x^2 ~dx\]
?

Answer 2

I get \[y=x^6/4+Cx^2\]
Solution claims\[2y=x^4+Cx^2\]

Answer 3

ohhhh hang on

Answer 4

ok i realied I wasnt mulitplying the right side by the integrating factor so now i get

\[y=x^4/2+Cx^2\]

Answer 5

gottt it thx

Answer 6

correction
\[\frac{ y }{ x^2 }=\int\limits x^3\times \frac{ 1 }{ x^2 }dx=\frac{ x^2 }{ 2 }+c\]
\[\frac{ 2 y }{ x^2 }=x^2+2c=x^2+ C\]
\[2 y=x^4+Cx^2\]
so you are correct.