1st order linear differential equation

[musicmar]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

integrating factor: e^{∫(x+2)/(x+1)}= e^xlx+1l

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

(y(e^xlx+1l)'

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

 

[hunt_mat]

First divide throughout to obtain an equation of the form:
<br /> \frac{dy}{dx}+P(x)y=Q(x)<br />
Then multiply through by the integrating factor and the LHS will be a total derivative, in your case it should be:
<br /> \left( e^{x}(1+x)y\right) &#039;=2x<br />

 

[musicmar]

Where did the 2x come from?

 

[HallsofIvy]

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]<br /> That is, we must have<br /> \frac{du}{dx}= \frac{x+2}{x+1}u<br /> or<br /> \frac{du}{u}= \frac{x+2}{x+1}dx= (1+ \frac{1}{x+2})dx<br /> <br /> Integrating both sides, ln(u)= x+ ln(x+2) so that<br /> u(x)= e^{x+ ln(x+2)}= (x+ 2)e^x<br /> <br /> What do you get when you multiply both sides of your equation by that?

 

[musicmar]

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

y'(x+2)e^x+[(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:

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

Taking the integral of both sides:

e^x(x+2)y=x²+c

y=(x²+c)/(e^x(x+2))


Yes?