Question

Find the Laplace transform Y(s)=L{y} of the solution of the given initial value problem.

y′′ +16y = {1, 0 ≤t < π
{0, π ≤ t < ∞, y(0) = 3, y′(0) = 5

Enclose numerators and denominators in parentheses. For example, (a−b)/(1+n).

Y(s) = ___________

Answer 1

$F(s)=L(y)=\int\limits_0^\infty e^{-s\cdot t}y(t)\,dt\\ L(y'')=\int\limits_0^\infty e^{-s\cdot t}y''(t)\,dt=0-y'(0)\cdot 1+s\cdot\int\limits_0^\infty e^{-s\cdot t}y'(t)\,dt=\\ =-5+s\cdot (0-y(0)\cdot 1+s\cdot \int\limits_0^\infty e^{-s\cdot t}y(t)\,dt)=\\ = -5-3\cdot s+s^2\cdot F(s).\\ \int\limits_0^\infty e^{-s\cdot t}f(t)\,dt=\int\limits_0^\pi e^{-s\cdot t}f(t)\,dt+\int\limits_\pi^\infty e^{-s\cdot t}f(t)\,dt=\\=\int\limits_0^\pi e^{-s\cdot t}\cdot 1\,dt+0=\frac{1-e^{-\pi s}}{s}.\\ -5-3\cdot s+s^2\cdot F(s)+16\cdot F(s)=\frac{1-e^{-\pi s}}{s}\\ (s^2+16)F(s)=5+3\cdot s+\frac{1-e^{-\pi s}}{s}=\frac{5s+3s^2+1-e^{-\pi s}}{s}\\ F(s)=\frac{5s+3s^2+1-e^{-\pi s}}{s(s^2+16)};$