Solve Linear ODE Simplify Step

[1s1]

Homework Statement

Solve: $x\frac{dy}{dx}-4y=x^{6}e^{x}$

Homework Equations

$x^{-4}\frac{dy}{dx}-4x^{-5}y=xe^{x}$ is equal to $\frac{d}{dx}[x^{-4}y]=xe^x$

The Attempt at a Solution

The second equation above simplifies to the third (according to my textbook) but I can't figure out how. Any help would be greatly appreciated!

 

[vela]

This website can display LaTeX. You just need to place the code between the appropriate tags. Take a look at https://www.physicsforums.com/showpost.php?p=3977517&postcount=3 and then fix your post.

 

[1s1]

Thanks!

 

[vela]

Look up "integrating factor." That's probably what the book is just about to show you how to find.

In this particular case, the first equation divided by x⁵ gives you the second equation, right?

 

[1s1]

Look up "integrating factor." That's probably what the book is just about to show you how to find.

In this particular case, the first equation divided by x⁵ gives you the second equation, right?

Yes, the second equation is just the first divided by x⁵. (Which is the first put in standard form by dividing through by x and then multiplying by the integrating factor which is x^-4 - so divided by x⁵ in total.)

The book then makes the jump that:

$x^{-4}\frac{dy}{dx}-4x^{-5}y=xe^{x}$ equals $\frac{d}{dx}[x^{-4}y]=xe^{x}$

and I can't figure out how this jump is made.

 

[LCKurtz]

The book then makes the jump that:

$x^{-4}\frac{dy}{dx}-4x^{-5}y=xe^{x}$ equals $\frac{d}{dx}[x^{-4}y]=xe^{x}$

and I can't figure out how this jump is made.

What do you get if you differentiate $x^{-4}y$ with respect to $x$ using the product rule?

 

[1s1]

What do you get if you differentiate $x^{-4}y$ with respect to $x$ using the product rule?

That's it! Thanks!