Question

(10 points) In this problem you will derive the formula for the Laplace transform of the second derivative of a function y. Use y and y′ for y(t) and y′(t), y0 and y1 for the initial conditions y(0) and y′(0), and Y for the Laplace transform of y. L{y′′(t)}(s)=∫[infinity]0 e−t y′′(t)dt

Answer 1

By definition of the Laplace transform,

$L(y''(t))=\displaystyle\int_0^\infty e^{-st}y''(t)\,\mathrm dt$

Integrate by parts:

$u=e^{-st}\implies\mathrm du=-se^{-st}\,\mathrm dt$

$\mathrm dv=y''(t)\,\mathrm dt\implies v=y'(t)$

$L(y''(t))=\displaystyle e^{-st}y'(t)\bigg|_0^\infty+s\int_0^\infty e^{-st}y'(t)\,\mathrm dt=sL(y'(t))-y'(0)$

Integrate by parts again to compute the transform of $y'(t)$:

$u=e^{-st}\implies\mathrm du=-se^{-st}\,\mathrm dt$

$\mathrm dv=y'(t)\,\mathrm dt\implies v=y(t)$

$L(y'(t))=e^{-st}y(t)\bigg|_0^\infty+\displaystyle s\int_0^\infty e^{-st}y(t)\,\mathrm dt=s L(y(t))-y(0)$

Putting everything together, we have

$L(y''(t))=s^2L(y(t))-sy(0)-y'(0)$

or in terms of the given initial conditions,

$\boxed{L(y''(t))=s^2L(y(t))-sy_0-y_1}$