ODE: Solving using Laplace Transform

1. Oct 16, 2016

RJLiberator

1. The problem statement, all variables and given/known data
Solve:
$$y''+λ^2y = cos(λt), y(0) = 1, y'(π/λ) = 1$$
where t > 0

2. Relevant equations

3. The attempt at a solution

I start off by taking the Laplace transform of both sides. I get:

$$L(y) = \frac{s}{(s^2+λ^2)^2}+\frac{sy(0)}{s^2+λ^2}+\frac{y'(0)} {s^2+λ^2}$$

Now take the inverse Laplace transform

$$y = \frac{1}{2λ}tsin(λt) + y(0)cos(λt)+\frac{y'(0)}{λ}sin(λt)$$

since y(0) = 1, we get

$$y = \frac{1}{2λ}tsin(λt) +cos(λt)+\frac{y'(0)}{λ}sin(λt)$$

Now this is where I get stuck. I have the boundary condition for y'(pi/λ)=1, but I'm not sure how to use it in this scenario.

If I take the derivative of y, then I get an expression with a y'(0) term in it that I'm not sure how to use...

Any thoughts?

2. Oct 16, 2016

Fightfish

Well, you end up with the solution of $y'(0)$ when you do that, no? If you put that back into your expression for $y(t)$, isn't that the solution to your DE?

3. Oct 16, 2016

vela

Staff Emeritus
Remember that $y'(0)$ is just a constant.

4. Oct 16, 2016

RJLiberator

Ah, beautiful, this was the hint I needed.

So I am capable of taking the derivative of y, since y'(0) is just some constant.

$$y'(t) = \frac{1}{2λ}[sin(λt)+λtcos(λt)]-λsin(λt)+y'(0)cos(λt)$$

Solving this for y'(π/λ) = 1 shows that y'(0) = -1-π/(2λ)

And thus, I have my determined y.