Second order differential equations

Question

Question:

Fine the continous and differentiable solution of the following equation:

y''(x) + 2y(x)=f (x)

Where:

f (x)={0 if x <0, 1 if 0 <x <1, 0 if 1 <x}

Satisfying the condition y (0)=0 and y'(0)=1.

I would know how to do the question if it was an actual function instead of f (x).

Any help would be much appreciated

Answer 1

You want to split your differential equation into homogenous and nonhomogenous parts, y=p+q.

The homogenous part is

p''+2p=0

I'm guessing this is the part you were talking about when you said you knew how to solve it. You find non-trivial solutions by assuming p=e^(Ax) and solving for A

The nonhomogenous part is

q''+2q=f(x)

Here, you need to intuitively figure out which value for q will give you f(x) in this equation. For the intervals where f(x)=0, you have q(x)=0 satisfying the equation. For the interval where f(x)=1, you'll see that q(x)=1/2 works out.

Find p and find q, then you get y=p+q