# Solving Initial Value Problems with Laplace Transforms

concept

All of our work up until now has been preparing us for this. We now finally have all of the skills needed to finally see a payoff in our study of the Laplace transform.
For a few of these examples you'll see that the Laplace transform is, in fact, a lot of work compared to other techniques we have.
Don't worry, there are many, many examples where the Laplace transform saves us considerable work, but those examples are very long and complex and so are not a good way learn the technique.
Just trust me that this is 100% worth your time, energy and effort despite it being hard to see at first.

fact

The process of solving an IVP with the Laplace transform is as follows:

- Take Laplace transform of equation
- Plug in the initial conditions
- Rearrange to get equation in form \(Y(s) = \)something
- Take inverse Laplace transform of both sides to get \(y(t) = \)something

The real trick to this method is in taking the inverse Laplace transform, which can be quite an exercise in algebra

example

Use the Laplace transform to solve the following I.V.P: \(y'' - y = t,\quad y(0) = 1,\quad y'(0) = 1\)

We first take the Laplace transform of everything. \(s^2Y(s) - sy(0) - y'(0) - Y(s) = \frac{1}{s^2}\) Now we plug in our initial conditions. \(s^2Y(s) - s - 1 - Y(s) = \frac{1}{s^2}\) Rearrange to get a formula for \(Y(s)\). \( Y(s)(s^2 - 1) = \frac{1}{s^2} + s + 1 \) \(Y(s) = \frac{\frac{1}{s^2} + s + 1}{s^2 - 1}\) \(Y(s) = \frac{s^3 + s^2 + 1}{s^2(s^2 - 1)} = \frac{s}{s^2 - 1} + \frac{1}{s^2 - 1} + \frac{1}{s^2(s^2 - 1)}\) Now we take the inverse Laplace of both sides. \(y(t) = \cosh(t) + \sinh(t) + \int_0^t\int_0^t \sinh(t) dtdt\) \(y(t) = \cosh(t) + 2\sinh(t)\)fact

If the conditions given with the differential equation are not at \(t=0\) we need to perform a substitution to get things into this form since the Laplace transform requires \(y(0)\) etc...
For some initial condition at \(t = t_0 > 0\):
Create a new function \(\tilde{y}(\tilde{t}) = y(t)\) where \(t = \tilde{t} + t_0\).
Now replace all your \(y(t)\)'s with \(\tilde{y}(\tilde{t})\)'s and do all of your usual IVP work. Then once you have your answer back in \(\tilde{t}\)'s we get things back in \(t\) using our above substitution.

example

Solve \(y'' + y = 2t,\quad y(\frac{1}{4}\pi) = \frac{1}{2}\pi,\quad y'(\frac{1}{4}\pi) = 2 - \sqrt{2}\pi\)

Our initial conditions aren't at \(t=0\) so we'll use the substitution \(t = \tilde{t} + \frac{1}{4}\pi\). We'll also use a new function \(\tilde{y}(\tilde{t}) = y(t)\). So our problem is now: \( \tilde{y}'' + \tilde{y} = 2(\tilde{t} + \frac{1}{4}\pi) \) If you're not sure how we got to this point keep going back over it until things click. Now we proceed as usual in solving this new problem. \( s^2\tilde{Y} - s\cdot \frac{1}{2}\pi - (2 - \sqrt{2}) + \tilde{Y} = \frac{2}{s^2} + \frac{\frac{1}{2}\pi}{s} + \frac{1}{2}\pi s + 2 - \sqrt{2} \) \( \tilde{Y} = \frac{2}{(s^2 + 1)s^2} + \frac{\frac{1}{2}\pi}{(s^2 + 1)s} + \frac{\frac{1}{2}\pi s}{s^2 + 1} + \frac{2 - \sqrt{2}}{s^2 + 1} \) Now we take the inverse term-by-term. We can see that the first two are the same as one of our above examples so: \( \tilde{y}(t) = 2(\tilde{t} - \sin(\tilde{t})) + \frac{1}{2}\pi (1 - \cos(\tilde{t})) + \frac{1}{2}\pi \cos \tilde{t} + (2 - \sqrt(2))\sin(\tilde{t}) \) \( 2\tilde{t} + \frac{1}{2}\pi - \sqrt{2}\sin(\tilde{t}) \) And finally we use our substitution \(t = \tilde{t} + \frac{1}{4}\pi \implies \tilde{t} = t - \frac{1}{4}\pi\) to get our answer back in terms of \(y(t)\) \( y(t) = 2t - \sin t + \cos t \)
practice problems