Solving an ODE with Laplace transforms

Question

One question, can you solve an ODE of type $y'''+y''\cdots =0$ with Laplace transforms if you don't get the values of $y_0$?

I've tried it but there are many unknowns.

Thanks

Answer 1

Assume that $y:[0,\infty)\to\mathbb R$ is $n+1$-times differentiable. We compute $$ \mathcal L(y'(t)) = \int_0^\infty e^{-st}y'(t)\ \mathsf dt = e^{-st}y(t)|_0^\infty + s\int_0^\infty e^{-st}y(t)\ \mathsf dt = Y(s) - y(0). $$ Assume now that $\mathcal L(y^{(n)}(t)) = s^n Y(s) - \sum_{j=1}^n y^{(n-j)}(0)s^{j-1}$ for some nonnegative integer $n$. Then \begin{align} \mathcal L(y^{(n+1)}(t)) &= \int_0^\infty e^{-st}y^{(n+1)}(t)\ \mathsf dt\\ &= e^{-st}y^{(n)}|_0^\infty + s\int_0^\infty e^{-st}y^{(n)}(t)\ \mathsf dt\\ &= y^{(n)}(0) + s\mathcal L(y^{(n)}(t))\\ &= y^{(n)}(0) + s(s^n Y(s) - \sum_{j=1}^n y^{(n-j)}(0)s^{j-1}\\ &= s^{n+1}Y(s) + y^{(n)}(0) - \sum_{j=1}^n y^{(n-j)}(0)s^j\\ &= S^{n+1}Y(s) + \sum_{j=1}^{n+1} y^{n+1-j}(0)s^{j-1}, \end{align} so by mathematical induction, the formula holds true for all nonnegative integers. Suppose $\sum_{i=0}^m a_iy^{(i)}(t) = 0$, then \begin{align} \mathcal L\left(\sum_{i=0}^m a_iy^{(i)}(t)\right) &= \sum_{i=0}^m a_i\mathcal L(y^{(i)}(t)\\ &= \sum_{i=0}^m a_i \left(s^i Y(s) - \sum_{j=1}^i y^{(i-j)}(0)s^{j-1}\right)\\ &= Y(s)\sum_{i=0}^m a_is^i - \sum_{i=0}^m\sum_{j=i} y^{(i-j)}(0)s^{j-1}\\ &= Y(s)\sum_{i=0}^m a_is^i - \sum_{j=1}^m\sum_{i=j}^m y^{(i-j)}(0)s^{j-1}\\ &= Y(s)\sum_{i=0}^m a_is^i - \sum_{j=1}^m s^{j-1} \sum_{i=0}^{m-j}y^{(i)}(0). \end{align} Now, since $\mathcal L\left(\sum_{i=0}^m a_iy^{(i)}(t)\right)=0$, we have $$ Y(s) = \frac{\sum_{j=1}^m s^{j-1} \sum_{i=0}^{m-j}y^{(i)}(0)}{\sum_{i=0}^m a_is^i}. $$ As you can see, we can compute the Laplace transform, but cannot solve the ODE without knowledge of the values of $y^{(i)}(0)$.