1st order linear differential equation

musicmar

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)}= 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:
[tex]
\frac{dy}{dx}+P(x)y=Q(x)
[/tex]
Then multiply through by the integrating factor and the LHS will be a total derivative, in your case it should be:
[tex]
\left( e^{x}(1+x)y\right) '=2x
[/tex]

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:
[tex]\frac{dy}{dx}+ \frac{x+2}{x+1}y= \frac{2x}{x+1}e^{-x}[/tex].
(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:
[tex]\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[/tex]
or
[tex]\frac{du}{u}= \frac{x+2}{x+1}dx= (1+ \frac{1}{x+2})dx[/tex]

Integrating both sides, [itex]ln(u)= x+ ln(x+2)[/itex] so that
[tex]u(x)= e^{x+ ln(x+2)}= (x+ 2)e^x[/tex]

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?