Using the Laplace Transform solve $y''+6y'+5y=e^t$

Question

The initial conditions are $y(0)=0$ and $y'(0)=1$. I began the process and ended up with $Y=1/(s-1)(s^2+6s+4)$. Since the second factor in the denominator does not factor so I have a feeling I messed something up along the way. Am I on the right track?

Answer 1

When taking the Laplace transform, note that \begin{align} \mathcal{L}\{\ddot{y}(t)\} &= s^2Y(s) - sy(0) - \dot{y}(0)\\ \mathcal{L}\{\dot{y}(t)\} &= sY(s) - y(0) \end{align} Therefore, the Laplace transform of your ODE the following \begin{align} s^2Y(s) - sy(0) - \dot{y}(0) + 6sY(s) - 6y(0) + 5Y(s) &= \int_0^{\infty}e^{t(1-s)}dt\\ Y(s)[s^2 + 6s + 5] -1 &= \frac{1}{s-1}\tag{1} \end{align} where we picked up the $-1$ from the initial condition $\dot{y}(0)=1$. We now have $$ Y(s) = \frac{1}{(s-1)(s^2+6s+5)}+\frac{1}{s^2+6s+5} $$ I suspect your mistake was with taking the Laplace transform of the $\ddot{y}(t)$ and $\dot{y}(t)$ or you assumed both initial conditions were zero. As for the $4$ instead of the $5$, I would guess that was just a typo.

Answer 2

A) Just because it doesn't factor doesn't mean you went wrong... however

B) you did go wrong because the denominator/bottom should have $s^2+6s+5$ rather than $s^2+6s+4$...

C) In fact, something else has gone awry because you should have

$$Y(s)=\frac{1}{(s-1)(s^2+6s+5)}+\frac{1}{s^2+6s+5}$$