First order non-separable linear deq's using an integration factor?

[QuarkCharmer]

For example:
\frac{dy}{dx} + y = e^{3x}
I understand that these differential equations are most easily solved by multiplying in a factor of integration, and then comparing the equation to the product rule to solve et al..

For example:

t\frac{dy}{dx} + 2t^{2}y = t^{2}
\frac{dy}{dx} + 2ty = t

Multiplying in an integration factor u(x), which in this case:
u(x) = e^{\int{2t}dt} = e^{t^{2}}

e^{t^{2}}\frac{dy}{dx} + 2te^{t^{2}}y = te^{t^{2}}

Now I can compress the left side down using the product rule and all that.

I don't understand how they are getting u(x) or why it's equal to e^{\int{2t}dt} ?

 

[micromass]

First we multiply the equation with an unkown function u(x)

u(x)\frac{dy}{dx}+u(x)y=u(x)e^{3x}

The goal is to recognize the left side as the derivative of a known function. Because then we can integrate both sides.

Now, remember the product rule

\frac{d}{dx}(u(x)y)=u(x)\frac{dy}{dt} + \frac{du(x)}{x}y

This looks a lot like the left side, doesn't it?? The only thing we want is

\frac{du(x)}{dx}=u(x)

Such a differential equation is easily solved. Indeed, it is clear u(x)=e^x satisfies this.

So now we got

\frac{d}{dx}{e^x y}=e^{4x}

Integrating both sides gives us

e^x y = \frac{1}{4}e^{4x}+C

or

y=Ce^{-x}+\frac{1}{4}e^{3x}

 

[QuarkCharmer]

Thanks Micro, this answers my questions perfectly.