# Using Laplace Transforms, find the solution to y′(t)+y(t−1)=t^2, with y(t)=0 for t <= 0.

Using Laplace Transforms, find the solution to ${y}^{\prime }\left(t\right)+y\left(t-1\right)={t}^{2}$, with y(t)=0 for $t\le 0$
You can still ask an expert for help

## Want to know more about Laplace transform?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Brooklynn Valencia
Performing the Laplace transform of the equation you obtain
$sY\left(s\right)-\underset{=0}{\underset{⏟}{y\left({0}^{-}\right)}}+{e}^{-s}Y\left(s\right)=\frac{2}{{s}^{3}}$
It follows
$Y\left(s\right)=\frac{2}{{s}^{3}}\frac{1}{s+{e}^{-s}}=\frac{2}{{s}^{4}}\frac{1}{1+\frac{1}{s}{e}^{-s}}=2\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{{s}^{n+4}}{e}^{-ns}$
It follows
$y\left(t\right)={\int }_{\eta -i\mathrm{\infty }}^{\eta +i\mathrm{\infty }}\frac{ds}{2\pi i}{e}^{st}Y\left(s\right)=2\sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}{\int }_{\eta -i\mathrm{\infty }}^{\eta +i\mathrm{\infty }}\frac{ds}{2\pi i}{e}^{st}\frac{1}{{s}^{n+4}}{e}^{-ns}$
$=\sum _{n=0}^{\mathrm{\infty }}\frac{2\left(-1{\right)}^{n}}{\left(n+3\right)!}\left(t-n{\right)}^{n+3}\mathrm{\Theta }\left(t-n\right).$
Where we've used Cauchy theorem in the last line.