I need to solve this differential equation using these conditions with the help of Laplace transform, but I am stuck. This question is from methods for solving differential equations. I am familiar on how to do Laplace transformations to an extent, however I am not sure how to apply it to the differential equation in question.
$$y'' + a^2y = f''(x);\quad \cases{y(0) = f(0)\cr y'(0) = f'(0)}$$
Here is my attempt:
We start by getting individual Laplace transforms of each one.
$\mathcal{L}\{y''\} = s^2Y(s) – sy(0) – y'(0)=s^2Y(s) – sf(0) – f'(0)$
$\mathcal{L}\{a^2y\} = a^2Y(s)$
$\mathcal{L}\{f''(x)\} = s^2 F(s) – sf(0) – f'(0).$
$$Y(s)=\frac{s^2}{s^2+a^2}F(s)$$
Then we try taking inverses and get
$$\mathcal{L}^{-1}\{Y(s)\}=s\cdot \frac{s}{s^2+a^2}\cdot F(s)=\delta'(t)*\text{cos}(at)*f(t)?$$
This is where I am stuck. By the way, are my calculations correct, because I highly doubt they are? How do I move further?
Best Answer
Given: \begin{align} y''(t) + a^2 \, y(t) &= f''(t) \\ y'(0) &= f'(0) \\ y(0) &= f(0) \end{align} and the modern standard Laplace transform $$ f(s) = \int_{0}^{\infty} e^{-s t} \, f(t) \, dt \quad \text{ or } \quad f(t) \doteqdot f(s) $$ then the follow can be determined.
Let $f''(t) = p(t)$, which for now is some function, then \begin{align} \mathcal{L}\{ y'' + a^2 \, y \} &= \mathcal{L}\{ p(t) \} \\ s^2 \, y(s) - y'(0) - s \, y(0) + a^2 \, y(s) &= p(s) \\ (s^2 + a^2) \, y(s) &= p(s) + y'(0) + s \, y(0) \\ y(s) &= \frac{p(s)}{s^2 + a^2} + \frac{y'(0)}{s^2 + a^2} + \frac{y(0) \, s}{s^2 + a^2} \\ &= \frac{1}{a} \, p(s) \, \frac{a}{s^2 + a^2} + \frac{y'(0)}{a} \, \frac{a}{s^2 + a^2} + y(0) \, \frac{s}{s^2 + a^2}. \end{align} The inverse of this result is $$ y(t) = \frac{1}{a} \, \int_{0}^{t} p(u) \, \sin(a(t-u)) \, du + \frac{y'(0)}{a} \, \sin(a t) + y(0) \, \cos(a t). $$
Now, returning to $p(t) = f''(t)$ leads to \begin{align} I &= \int_{0}^{t} p(u) \, \sin(a(t-u)) \, du \\ &= \int_{0}^{t} f''(u) \, \sin(a(t-u)) \, du \\ &= a \, f(t) - f'(0) \, \sin(a t) - a \, f(0) \, \cos(a t) - a^2 \, \int_{0}^{t} f(u) \, \sin(a(t-u)) \, du \end{align} which gives $y(t)$ as: \begin{align} y(t) &= f(t) + \frac{y'(0) - f'(0)}{a} \, \sin(a t) + (y(0) - f(0)) \, \cos(a t) \\ & \hspace{15mm} - a \, \int_{0}^{t} f(u) \, \sin(a(t-u)) \, du. \end{align} Using $y'(0) = f'(0)$ and $y(0) = f(0)$ reduces the result to $$ y(t) = f(t) - a \, \int_{0}^{t} f(u) \, \sin(a(t-u)) \, du. $$
Notes:
The problem proposer has obtained the result $$ Y(s) = \frac{s^2 \, F(s)}{s^2 + a^2}. $$ In the method shown to obtain this result the conditions given were already used. What is left is just to determine the inverse of the transform. This can be done by considering: \begin{align} Y(s) &= \frac{s^2 \, F(s)}{s^2 + a^2} \\ &= \left( 1 - \frac{a^2}{s^2 + a^2} \right) \, F(s) \\ &= F(s) - a \, \frac{a}{s^2 + a^2} \, F(s) \end{align} which has the inverse $$ y(t) = f(t) - a \, \int_{0}^{t} \sin(a(t-u)) \, f(u) \, du. $$
This gives the same result in both demonstrations (the proposer's work by using the conditions first and the solution presented which uses the conditions last).