1. ## 2nd ODEs

Having a few problems

1. $d^2y/dx^2 + 4dy/dx + 5y= x + 2$

So the characteristic equation is $r^2+4r+5=0$
with imaginary roots so

$y(h)= e^-2x(C(1)cosx+ C(2)sinx)$

But how do i solve for y(p)? I cant do 5y=x+2.. thats not working

I think i'll be fine for all of them once I figure out how to solve these when there is an x variable on the right side and not just a 0 or a constant

2. The solution to an inhomogenous linear equation is the sum of the general solution to the homogenous equation plus a particular solution. Its not easy in general to find a particular solution; most people just guess and check I think. In your case, try guessing $y_{\text{particular}}(x) = bx+c$ and solve for b and c. Then the general solution will be $y =e^{-2x}(C_1\cos x+ C_2\sin x) + bx + c$.

3. Are there any strategies other than just guessing?

4. In general? Not that I am aware of. If the RHS is a polynomial and the LHS has constant coefficients, the particular soluiton will always be a polynomial of the same degree however.

5. Originally Posted by Jeffman50
Are there any strategies other than just guessing?
http://www.mathhelpforum.com/math-he...-tutorial.html

See posts #6,7,8