Question

ODE

xy'+(2x-3)y=4x^4


After I get to e^2x x^-3 I get lost.

Answer 1

separable
xy'+(2x-3)=4x^4
\[y'=\frac{4x^4-2x+3}{x} \\ \\ \int\limits \frac{4x^4-2x+3}{x} dx\]
\[x^4-2x+3\ln(x)+c\]

Answer 2

I wrote it wrong, it's xy'+(2x-3)y=4x^4

Answer 3

after finding the integrating factor, which is \[e^{2x} x^{-3}\] you now solve the equation
\[(e^{2x}x^{-3})y=\int(e^{2x}x^{-3})(4x^4)dx\]

Answer 4

in the ODE \[y'+P(x)y=Q(x)\]
the integrating factor is \[v=e^{\int P(x) dx}\]
and the solution to the ODE is given by
\[vy=\int v\cdot Q(x) dx\]

Answer 5

Thanks!