# Simple differential equation

1. Jul 22, 2014

### johann1301

1. The problem statement, all variables and given/known data

a) Write (x21)y'+2xy as the derivative of a product
b) Solve (x21)y'+2xy=e-x

3. The attempt at a solution

a) I use the product rule backwards and get

((x2+1)y)'

b) I exploit what i just found out...

(x21)y'+2xy=((x2+1)y)'

and get...

e-x=((x2+1)y)'

integrate on both sides...

∫e-xdx=∫((x2+1)y)'dx

-e-x+C=(x2+1)y

and get that...

y=(C-e-x)(x2+1)-1

this is the correct answer according to the book.

What i am curious about is in the step marked with bold text. I wrote dx at the end just by guessing. Why couldnt i just have written dy instead?

2. Jul 22, 2014

### slider142

You could have integrated both sides with respect to y if you really wanted to, but while the left side would then become ye-x, there is no theorem that helps us easily integrate the right side with respect to y. We are only able to integrate it with respect to x because we found in step one that it is the derivative with respect to x of (x2 + 1)y, and the fundamental theorem of calculus then assures us that its integral with respect to x is (x2 + 1)y plus an arbitrary constant.

3. Jul 22, 2014

### johann1301

Thank you!

4. Jul 22, 2014

### ehild

y is function of x and the comma ' means derivative with respect to x. y ' means dy/dx. If f=(x2+1)y then
f '= df/dx . You get f if you integrate f '. ∫f ' dx= f

Formally you can handle the problem as if df/dx was a simple fraction. df/dx = e-x, multiply both sides by dx df= e-x dx and put the integral symbol at the front:

∫ (df/dx) dx = ∫e-x dx

ehild

5. Jul 23, 2014

### HallsofIvy

Staff Emeritus
You mean (x2+ 1)y'. That puzzled me for a while!

6. Jul 23, 2014

### johann1301

Sorry!

7. Jul 23, 2014

### johann1301

This is only true in this case right? Its more just an assumtion we make based on the look of the task?

8. Jul 23, 2014

### slider142

Yes. In most texts on ordinary differential equations, y is assumed to be the dependent variable and x is assumed to be the independent variable as a matter of notation only, usually established by the author in the first chapter. Therefore y' is assumed to always mean y'(x), or dy/dx in these texts only.

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