# Simple differential equation

## Homework Statement

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

## 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?

## Answers and Replies

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.

1 person
Thank you!

ehild
Homework Helper
...............
e-x=((x2+1)y)'

integrate on both sides...

∫e-xdx=∫((x2+1)y)'dx

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

..................

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?

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

HallsofIvy
Homework Helper

## Homework Statement

a) Write (x21)y'+2xy as the derivative of a product
b) Solve (x21)y'+2xy=e-x
You mean (x2+ 1)y'. That puzzled me for a while!

## 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?

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

Sorry!

the comma ' means derivative with respect to x

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

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

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.