Printable View
I can't get this one, and it's annoying me.
Find the general solution to the following differential equation, and the corresponding particular solution for the initial value given. Primes denote derivatives with respect to x.
[q](x2 + 4)y' + 3xy = x; y(0) = 1[/q]
Start by changing your notation to dy/dx instead of primes or dots. This turns the equation into (x2+4)*(dy/dx) +3xy = x. Since it's a first order diff eq we have to solve it like a first order. Either we can do it as though it were separable, or through integrating factors.
Let's be little bitches and assume that's it's separable. Integrating factors suck.
Multiply both sides so that you don't have to deal with derivatives in denominators:
(x2+4)*dy + 3xy(dx) = x*dx
Bring the 3xy to the other side:
(x2+4)*dy = x*dx - 3xy(dx)
Notice that we have a common term on the right side: x*dx, so factor it out. This brings us closer to getting y and dy terms on one side, and x and dx terms on the other.
(x2+4)*dy = x*dx*(1-3y)
Ohh look, a y term by itself, meaning that it was separable. Bring the x2+4 to the other side with algebra:
dy/(1-3y) = x*dx/(x2+4)
Now, kill it with integrals. You should have a natural log on both sides. Recall back to Calculus: you'll need u subs.
What to do with y(0) = 1? Well, that's our initial condition. Just find an equation where the equation goes through (0,1). That should be a quick look through our notes on how to deal with these.