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

1. Jan 22, 2012

### 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} ?

Last edited: Jan 22, 2012
2. Jan 22, 2012

### 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

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

Last edited: Jan 22, 2012
3. Jan 23, 2012

### QuarkCharmer

Thanks Micro, this answers my questions perfectly.