Solving xy' - 4y = x⁴eˣ with integrating factor

[BarackObama]

Homework Statement

xy' - 4y = x⁴e^x

Homework Equations

The Attempt at a Solution

y' - 4x^-1y = x³e^x
x^-4y' - 4x^-5y = x^-1e^x

I'm not sure what to do next, I can't express the LS as a derivative

 

[Mark44]

Homework Statement

xy' - 4y = x⁴e^x

Homework Equations

The Attempt at a Solution

y' - 4x^-1y = x³e^x
x^-4y' - 4x^-5y = x^-1e^x

I'm not sure what to do next, I can't express the LS as a derivative

The whole purpose of finding an integrating factor is so that you can rewrite the left side as the derivative of something.

In this case, the left side is the derivative of x^-4y, which you can easily verify.

 

[PAllen]

So, am I to make anything out of BarackObama asking about an integrating factor?

Try letting y=(x^4)u , see if u is easier to solve for.

 

[HallsofIvy]

Homework Statement

xy' - 4y = x⁴e^x

Homework Equations

The Attempt at a Solution

y' - 4x^-1y = x³e^x
x^-4y' - 4x^-5y = x^-1e^x

I'm not sure what to do next, I can't express the LS as a derivative

The whole point of an "integrating factor", u, is that the left side becomes (uy)'. You found $x^{-4}$ as integrating factor so the left side must be $(x^{-4}y)'$.

And, in fact, by the product rule, $(x^{-4}y)'= x^{-4}y'- 4x^{-5}y$.