Question

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

y′′ + 4y = { t, 0 ≤ t < 1
{ 1, 1 ≤ t < ∞, y(0) = 3, y′(0) = 3


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

Y(s)= _______________

Answer 1

$Y(s)=L(y)=\int\limits_0^\infty e^{-st}y(t)\,dt.\\ L(y'')=-3-3\cdot s + s^2\cdot \int\limits_0^\infty e^{-st}y(t)\,dt=\\ =-3-3\cdot s + s^2\cdot Y(s).\\ L(f(t))=\int\limits_0^\infty e^{-st}f(t)\,dt=\int\limits_0^1 e^{-st}f(t)\,dt+\int\limits_1^\infty e^{-st}f(t)\,dt=\\ =\int\limits_0^1 e^{-st}t\,dt+\int\limits_1^\infty e^{-st}\,dt=\frac{1-e^{-s}-se^{-s}}{s^2}+ \frac{e^{-s}}{s}.\\\\ -3-3\cdot s + s^2\cdot Y(s)+4\cdot Y(s)=\frac{1-e^{-s}-se^{-s}}{s^2}+ \frac{e^{-s}}{s}\\ (s^2+4)\cdot Y(s)=3+3\cdot s+\frac{1-e^{-s}-se^{-s}}{s^2}+ \frac{e^{-s}}{s}\\ Y(s)=\frac{3+3\cdot s+\frac{1-e^{-s}-se^{-s}}{s^2}+ \frac{e^{-s}}{s}}{s^2+4};$