Easy algebra question related to Differential equations

[eugenius]

I have this really simple differential equation. It needs to be in the form y'+ p(x)y=q(x) to solve it as a linear first order ODE. I am almost embarassed to ask, but I can't seem to get it in exactly this form. Read below.

Homework Statement

y'+3y=2x(e^-3x)



Need y'+ p(x)y=q(x)

The Attempt at a Solution

Divide both sides by 2x, gives

y'/2x + 3y/2x = e^-3x

Now here I have y' multiplied by the function 1/2x, but according to the proper linear first order ODE form, Y' needs to be by itself.

Can you give me some tips on how to achieve this? I don't know why I can't see it, its probably very simple, but whatever the algebraic manipulation is for this, I seem to have forgotten it.

Thank you.

 

[Defennder]

It's already in that required form. You have y' + 3y =2xe^(-3x). RHS is a function of x, and 3 is p(x). Remember that a constant function is a function of any variable.

 

[eugenius]

Ah so p(x)= 3. Yes I see. Thanks.