Initial value problem using Laplace Transform

Question

I have to solve the IVP: $$\begin{cases} y''+by'-cy=1\\ y(0)=y_0\\ y'(0)=y'_0 \end{cases}$$

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

We have to find $b,c,y_0$ and $y'_0$

I have found that: $$(s^2+s+1)Y(s)-(s+1)y_0-y'_0=1/s$$

How do I proceed further

Answer 1

According to the given ODE $y''+by'-cy=1$, your last line should be $$(s^2+bs-c)Y(s)-(s+b)y_0-y'_0=1/s$$ which implies $$Y(s)=\frac{y_0s^2+(by_0+y'_0)s+1}{s^3+bs^2-cs}.$$ Now, by comparing it with $Y(s)=\frac{s^2+2s+1}{s^3+3s^2+2s}$, we are able to find $b$, $c$, $y_0$ and $y'_0$ easily: $$b=3,c=-2, y_0=1, y'_0=-1.$$

Answer 2

You have that the Laplace transform of the solution $y(t) $ is $Y(s) $. So the inverse Laplace transform of $Y(s) $ gives us the solution $y(t) =\frac{e^{-2t}+1}{2}$. From this we can calculate $y(0)=1$ and $y'(0) =-1$. Now sustitute the expression of $y(t)$ in the equation and you'll get $$e^{-2t}(4-2b+c) -(2+c) =0.$$ Since $e^{-2t}$ and $1$ are linearly independent we Have that $b=3$ and $c=-2$.