Solve the initial value problem via convolution of Laplace transforms

Question

$y''-y=\sin2t+1, y(0) = 1, y'(0) = -1$

Hey all, I am not sure if I am completing all the steps correctly to solve this problem via convolution. I am pretty sure I am not doing it correctly, but am not sure what is going wrong.

First I apply Laplace transforms to everything. $$\mathcal{L}[y''] - \mathcal{L}[y] = \mathcal{L}[\sin2t] + \mathcal{L}[1]$$

Plugging in initial values and doing algebra gives me $$Y(s) = \left(\frac{1}{s^2+4} + \frac{1}{s} \right) \left(\frac{1}{s+1} \right)$$

Now I let

\begin{align*} \mathcal{L}[f](s) &= \left(\frac{1}{s^2+4} + \frac{1}{s} \right) \\ \mathcal{L}[g](s) &= \left(\frac{1}{s+1} \right) \end{align*}

I get $f(t) = \sin2t+1$ and $g(t) = e^{-t}$ via inverse laplace transform.

Now that I have $f$ and $g$, I want to do a convolution $f*g = \int_{0}^{t} f(x)g(t-x)dx$ , but I am unable to solve the integral so I think I have done something wrong before getting to this step.

Answer 1

You computed your initial Laplace transform incorrectly. Notice that

\begin{eqnarray*} \mathcal{L}[y''] - \mathcal{L}[y] & = & s^2\mathcal{L}[y] - y(0) - y'(0) - \mathcal{L}[y] \\ & = & (s^2-1)\mathcal{L}[y] - 1 - (-1) \;\; =\;\; (s^2-1)\mathcal{L}[y]. \end{eqnarray*}

You divided by the wrong term on the right hand side. You should've obtained

$$ Y(s) \;\; =\;\; \left (\frac{1}{s^2+4} + \frac{1}{s}\right )\left (\frac{1}{s^2 - 1}\right ) $$

hence $f(t) = \frac{1}{2}\sin(2t) + 1$. Because $\frac{1}{s^2-1} = \frac{1}{(s+1)(s-1)} = \frac{1}{2(s+1)} - \frac{1}{2(s-1)}$, then we find that $g(t) = \frac{1}{2}e^{-t} - \frac{1}{2}e^t$.

Answer 2

$$y''-y=\sin2t+1, y(0) = 1, y'(0) = -1$$ I got this: $$Y(s) = \left(\frac{2}{s^2+4} + \frac{1}{s} \right) \left(\frac{1}{s^2-1} \right)+\dfrac 1 {s+1}$$ Which is decomposed as: $$Y(s) = -\frac{2}{5(s^2+4)} + \frac 1{5(s-1)} +\frac{s}{s^2-1} - \frac{1}{s} +\dfrac 4 {5(s+1)}$$ $$y(t)=-1-\frac 15 \sin(2t)+\frac 15e^t+\cosh(t)+\frac 45e^{-t}$$ Finally: $$\boxed {y(t)=-1-\frac 15 \sin(2t)+\frac {7}{10}e^t+\frac {13}{10}e^{-t}}$$

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Edit: I didn't see that OP need to use Convolution Integral; $$Y(s) = \left(\frac{2}{s^2+4} + \frac{1}{s} \right) \left(\frac{1}{s^2-1} \right)+\dfrac 1 {s+1}$$ $$y(t) = \int_0^t \sinh (\tau)\sin(2(t-\tau))d\tau +\int_0^t \sinh (t-\tau)d\tau +e^{-t}$$ $$y(t) = \int_0^t \sinh (\tau)\sin(2(t-\tau))d\tau +\cosh(t)-1 +e^{-t}$$ Integrate by part for the first integral. $$I=\int_0^t \sinh (\tau)\sin(2(t-\tau))d\tau$$ $$I=\frac 25\sinh t -\frac 15\sin(2 t)$$ Finally : $$ y(t) = \frac 25\sinh t -\frac 15\sin (2t) +\cosh(t)-1 +e^{-t}$$ $$\boxed {y(t)=-1-\frac 15 \sin(2t)+\frac {7}{10}e^t+\frac {13}{10}e^{-t}}$$