Second Order Differential Eqn.

http://www.mediafire.com/view/?grhasyd8ilzemsw


I'm not completely sure what I'm being asked to do in the first part.

This is what I've done :

I let y = e^(mx) and the substitute into the given equation to get a quadratic:

m²-bm+4 Since b<0 it must be negative

I then complete the square to obtain two solutions :

Click here

The problem that I now face is how do I express the general solution. Depending on the value of b^2 in the square root I may have something of the form:

y=Acoswx + Bsinwx

or

y=Ae^x+Be^x

The critical case is when the quadratic has a repeated root. For which value negative of b does this occur? You should also know that when the root is repeated (call it r), the solution to the second-order linear homogenous ODE is (A+Bx)exp(rx).

Ah, I see. I never knew that referred to a repeated root case. So when b^2 = 4ac b^2 = 16 Hence, my answer to the quadratic simplifies to :

(16)^0.5 / 2 = ±2 However, we are told b<0 hence the general solution is :

y = (A+Bx)e^2x

Is that correct ?

Thank you everyone ! +rep to all.