Question

Solve the DE using Laplace Transforms y''+9y=0 y(0)=0 y'(0)=2

Answer 1

So far I have gotten to this:

\[L(y)=2 \div (s^2+9)\]

Answer 2

do you have to use Laplace transforms

Answer 3

Hmm ok your work so far looks good.
Let's multiply the top and bottom by 3,\[\large \mathscr{L}\left[y\right]=\frac{3}{3}\cdot\frac{2}{s^2+9}\]Which will give us,\[\large \mathscr{L}\left[y\right]=\frac{2}{3}\cdot\color{green}{\frac{3}{s^2+3^2}}\]

Now you're able to take the inverse laplace transform.

Answer 4

That green term should look like a familiar trig laplace transform

Answer 5

@zepdrix Thanks, but what about the initial conditions? I don't really see any constants to solve for.

Answer 6

I got y(t)=2/3 sin (3t)