Solving ODE with laplace transform question
Hi,
I've been trying to work on this problem for a bit now to solve the ODE: $5y''+2y'+y=1$ , where $y'(0)=1$ and $y(0)=0$. But I guess I'm struggling with the algebra or the steps somewhere. I have reached the point where I get $\frac{(1+5s)}{(s(5(s+0.2)^2+0.4^2))}$ then I get very lost.
The answer my lecturer gave is ${[Y(s)]}=1-e^{-0.2t}\cos(0.4t)+2e^{-0.2t}\sin(0.4t)$
Any help is extremely appreciated!!
Edit:
So I have reached the point where I have done partial fractions decomposition and I got $\frac{1}{s} + \frac{(0.6-s)}{((s+0.2)^2+0.4^2)}$ so that would give me $1-e^{-0.6t}cos(0.4t)$. Do I have to separate something again to get the correct answer? Or is this an acceptable answer?
Edit 2:
Thanks to the comments, I realized I had incorrectly used a table. I wanted to use the Laplace rule: $e^{\beta t}cos(\alpha t) = \frac{s-\beta}{(s-\beta)^2+\alpha^2}$.
But to use this, the $\beta$ in the numerator and denominator need to be the same so I needed to use algebra to separate these and then use the laplace rule: $e^{\beta t}sin(\alpha t) = \frac{\alpha}{(s-\beta)^2+\alpha^2}$.
Thank you for all the help!!
As far as can be seen, somewhere on paper you have the right result up to nd including the partial fraction decomposition. However in the inverse transform stage there are some grievous errors. Check again your sources on the proper procedure.
If you have no table at hand, the rules you need to apply are \begin{align} {\cal L}^{-1}\left\{Y(s+a)\right\}(t) &=e^{-at}{\cal L}^{-1}\left\{Y(s)\right\}(t) \\ {\cal L}^{-1}\left\{\frac{a}{s^2+a^2}\right\}(t) &=\sin(at) \\ {\cal L}^{-1}\left\{\frac{s}{s^2+a^2}\right\}(t) &=\cos(at) \end{align}