Science:Math Exam Resources/Courses/MATH215/December 2013/Question 04 (b)/Solution 1

We take the Laplace transform of the equation:

${\mathcal {L}}(y''-2y'+y)={\mathcal {L}}(g(t))$

$\underbrace {s^{2}Y(s)-sy(0)-y'(0)} _{{\mathcal {L}}(y'')}-2\underbrace {(sY(s)-y(0))} _{{\mathcal {L}}(y')}+\underbrace {Y(s)} _{{\mathcal {L}}(y)}=\underbrace {\frac {e^{-s}}{s-1}} _{{\mathcal {L}}(g(t))}$

$s^{2}Y(s)-0-1-2(sY(s)-0)+Y(s)=(s^{2}-2s+1)Y(s)-1={\frac {e^{-s}}{s-1}}$

Since $\displaystyle s^{2}-2s+1=(s-1)^{2}$ we arrive at our final answer

$Y(s)={\frac {e^{-s}}{(s-1)^{3}}}+{\frac {1}{(s-1)^{2}}}$