Why Can't I Use dy Instead of dx in Solving This Differential Equation?

[johann1301]

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 couldn't i just have written dy instead?

 

[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.

 

[johann1301]

Thank you!

 

[ehild]

...
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 couldn't 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 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 couldn't i just have written dy instead?

 

[johann1301]

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

Sorry!

 

[johann1301]

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?

 

[slider142]

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.