Question

Solve the DE using Laplace Transforms

y''+9y=0
y(0)=0
y'(0)=2

Answer 1

We need a bit of definition to run through this, primarily the Laplace transform of a derivative.
\( \large \displaystyle \color{gold}{*} \; \mathcal{L} (y") = s^2 \mathcal{L}(y) - s \; y(0) - y'(0) \)

Now, we start with our DE:
\( \displaystyle y" + 9y = 0 \)
We take the Laplace transform of both sides, thus preserving equality.
\( \displaystyle \mathcal{L} (y" + 9y) = \mathcal{L} (0) \)
Laplace transform has the property of linearity (conserved from distribution and linearity of integration)
\( \displaystyle \mathcal{L} (y") + 9 \mathcal{L} (y) = 0 \)
Now, we have our definition for the Laplace transform of the derivative.
\( \displaystyle \color{#999955}{s^2 \mathcal{L} (y) - s \; y(0) - y'(0)} + 9 \mathcal{L} (y) = 0 \)

We have like-terms on the left-hand side, which we can factor out common terms.
\( \displaystyle (s^2 + 9) \mathcal{L} (y) - s \; y(0) - y'(0) = 0 \)
We may now use our initial conditions: \( y(0) = 0, y'(0) = 2 \)
\( \displaystyle (s^2 + 9) \mathcal{L} (y) - s \times \color{#999955}{0} - \color{#999955}{2} = 0 \)
\( \displaystyle (s^2 + 9) \mathcal{L} (y) - 2 = 0 \)
It is just some manipulations to isolate our Laplace transform.
\( \displaystyle (s^2 + 9) \mathcal{L} (y) = 2 \)
\( \displaystyle \mathcal{L} (y) = \frac{2}{s^2 + 9} \)
For our solution, we simply take the inverse Laplace transform of each side now.
\( \displaystyle y = \mathcal{L}^{-1} \left(\frac{2}{s^2 + 9} \right) \)
\( \displaystyle y = 2 \mathcal{L}^{-1} \left(\frac{1}{s^2 + 9} \right) \)

If you are familiar with the various standard Laplace transforms and their inverses, you might recognise what that Laplace transform comes out to be. Otherwise, we can use a table to figure it out.
\( \large \displaystyle \color{gold}{*} \; \mathcal{L}^{-1} \left(\frac{a}{s^2 + a^2}\right) = sin (at) \)
So, we can fit that description by multiplying with a magic one: 3/3
\( \displaystyle y = \frac{2}{3} \mathcal{L}^{-1} \left( \frac{3}{s^2 + 3^2} \right) \)
\( \displaystyle y = \frac{2}{3} \sin 3t \)

And there we have it! We are done!