How to determine general solution using Laplace transform?

Question

$y′′ + a^2y = 2u(t-10)$

Here $a > 0$ and is any real number. I am confused by $a^2$ value there. Can anyone show me step by step how to get to up to $Y(s)$? It would really help me understand. Thank you.

Answer 1

If you take Laplace transform:

$$s^2Y(s)-sy(0)-y'(0)+a^2Y(s)=\frac{2e^{-10s}}{s}.$$

You can verify that $\mathscr L\{y''\}=s^2Y(s)-sy(0)-y'(0)$, and that $\mathscr L\{2\,\mathcal U(t-10)\}=\dfrac{2e^{-10s}}{s}$ using the definition $\mathscr L\{f(t)\}=\int_0^\infty e^{-st}f(t)\,dt$.

For the first part:

$$\mathscr L\{y''\}=\int_0^\infty e^{-st}y''(t)dt=e^{-st}y'(0)\Big|_0^\infty- s\int_0^\infty e^{-st}y'(t)\,dt=-y'(0)+s\mathscr L\{y'\}.$$ You can take it from here, the Heaviside step function can be done easily as well.