Question

Can someone show the solution to this DE?

x*dy/dx-4y = (e^x)*(x^4)

Answer 1

divide by x and you will have an equation in the form \[\frac{ dy }{ dx } - \frac{4}{x}y = (e^{x})(x^{4})\frac{1}{x}\]

Answer 2

do you know how what an integrating factor is?

Answer 3

yeah, i tried it, i got all messed up. can you show me?

Answer 4

what did you have as the integrating factor?

Answer 5

4/x and x^3 * e^x

Answer 6

so the integrating factor is \[p = \exp(\int\limits_{}^{} g(x) dx)\]

Answer 7

where in your case \[g(x) = \frac{-4}{x}\]

Answer 8

\[\frac{dy}{dx} + g(x)y = h(x)\]

\[p = \exp(\int\limits\limits_{}^{} g(x) dx)\] where p is the integrating factor. times everything through by p

\[p\frac{dy}{dx} + pg(x)y = ph(x)\]

now you can rewrite the left hand side because:
\[\frac{d}{dx}(p(x)y) = p\frac{dy}{dx} +pg(x)y\]

Answer 9

and finally integrating gives

\[p(x)y = \int\limits_{}^{}p(x)h(x)dx\]

Answer 10

divide by p(x) to give y in terms of an integral. Integrate and you have solved for y