# 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!

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1s1
Thanks!

Staff Emeritus
Homework Helper
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?

Last edited:
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.

Homework Helper
Gold Member
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!