Separation of variables to solve DE

[beetle2]

Homework Statement

y' + (2/x)y = 3/x^2

Homework Equations

separation of variables

The Attempt at a Solution

First I turned it into

dy/dx + (2/x)y = 3/x^2 dx


then multiplied both sides by dx

dy + (2/x)y = 3/x^2 dx

I then tried to divide both sides by 2/x and got

dy + y = 3/2x

Do I just integrate both sides now?

 

[Bohrok]

Separation of variables won't help (you also made an error when multiplying through the equation with dx). It looks like you'll have to solve this using an integrating factor.

 

[Mindscrape]

You also made an error when dividing by (2/x). Make sure that what you do to one term goes for all terms.

 

[boneill3]

The integrating factor is $e^{\int{\frac{2}{x}dx}$ = $x^2$

We than use this and multiply both sides by $x^2$

which gives


$x^2y'+2xy = 3$

or

$yx^2= 3x+C$

divide both sides by $x^2$

$y = \frac{3}{x}+\frac{C}{x^2}$

 

[beetle2]

Is this the general solution?

 

[Raskolnikov]

yes, it's the general solution.

 

[gomunkul51]

Also can be solved as a non-homogeneous Euler equation.

 

[beetle2]

which gives


$x^2y'+2xy = 3$

or

$yx^2= 3x+C$

I see you integrated on the right side wrt x to get 3x+C

but how did $x^2y'+2xy$ become $yx^2$? did we integrate the LHS wrt x or y?

 

[Raskolnikov]

Given a function $$f(x,y)$$, its total differential with respect to x is $$\frac{df}{dx} = \frac{\partial f}{\partial x}\frac{dx}{dx} + \frac{\partial f}{\partial y}\frac{dy}{dx}.$$

 

[vela]

I see you integrated on the right side wrt x to get 3x+C

but how did $x^2y'+2xy$ become $yx^2$? did we integrate the LHS wrt x or y?

You have to treat both sides of the equation the same way. If you integrate the RHS wrt x, you have to integrate the LHS wrt x as well.