## 1st order linear differential equation

1. The problem statement, all variables and given/known data
I'm trying to study for a quiz tomorrow by doing some practice problems. If someone could help me with the process of solving a 1st order linear diff. eq., that would be great.

(x+1)(dy/dx) + (x+2)y = 2xe-x

2. Relevant equations

3. The attempt at a solution

dy/dx + [(x+2)/(x+1)]y = 2xe-x/(x+1)

integrating factor: e∫(x+2)/(x+1)= exlx+1l

This is where I get confused. I should be able to use the product rule here:

(y(exlx+1l)'

so that I will be able to take the integral of (above) and {2xe-x/(x+1)]*[exlx+1l].

Once I take the integrals, then I can solve for c(not in this problem, though) and try to solve for y explicitly. Some help with the middle steps would be greatly appreciated.
 Recognitions: Homework Help First divide throughout to obtain an equation of the form: $$\frac{dy}{dx}+P(x)y=Q(x)$$ Then multiply through by the integrating factor and the LHS will be a total derivative, in your case it should be: $$\left( e^{x}(1+x)y\right) '=2x$$
 Where did the 2x come from?

Recognitions:
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## 1st order linear differential equation

The point was that the formula you used for the integrating factor requires that the coefficient of the derivative be 1. Here it is x+ 1 so you need to divide the entire equation by x+1:
$$\frac{dy}{dx}+ \frac{x+2}{x+1}y= \frac{2x}{x+1}e^{-x}$$.
(The first equality is from the product rule, the second from just multiplying the left side of the differential equation by u.)

Now, you are looking for a function, u(x), so that multiplying by it will make that left side a single derivative:
$$\frac{d(u(x)y)}{dx}= u(x)\frac{dy}{dx}+ \frac{du}{dx}y= u\frac{dy}{dx}+ \frac{x+2}{x+1}u y[/itex] That is, we must have [tex]\frac{du}{dx}= \frac{x+2}{x+1}u$$
or
$$\frac{du}{u}= \frac{x+2}{x+1}dx= (1+ \frac{1}{x+2})dx$$

Integrating both sides, $ln(u)= x+ ln(x+2)$ so that
$$u(x)= e^{x+ ln(x+2)}= (x+ 2)e^x$$

What do you get when you multiply both sides of your equation by that?
 dy/dx + [(x+2)/(x+1)]y = 2xe-x/(x+1) y'(x+2)ex+[(x+2)2/(x+1)]y=2x Ok, now I see where the 2x comes from. Do I have the rest right? If so, then by the product rule I should have: (ex(x+2)y)'=2x Taking the integral of both sides: ex(x+2)y=x2+c y=(x2+c)/(ex(x+2)) Yes?
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