Find $y(t)$ in a causal system with input-output relationship
$$\frac{dy(t)}{dt} + 3y(t) = x(t)$$
where
$$x(t) = e^{2t} \cdot u(-t).$$
and $u$ is the Heaviside function.
To try and solve this problem, I first take the Laplace transform of both sides to get
$$sY(s) - y(0) + 3Y(s) = 0.$$
This can be simplified to
$$Y(s)(s + 3) - y(0) = 0,$$
which gets me $$Y(s) = \frac{y(0)}{s + 3}$$ So, I find $y(t) = e^{-3t} u(t) \cdot y(0)$. Is this correct?
I have a Laplace transform table that tells me the Laplace transform of $-e^{-\alpha t} u(-t) = \frac{1}{s + \alpha}$. Why does this not work here?
When $t < 0$, I took the Laplace transform of both sides to get
$$sY(s) - y(0) + 3Y(s) = \frac{1}{s-2}.$$
So, I find
$$Y(s)(s + 3) - y(0) = \frac{1}{s - 2},$$
from which I get
$$Y(s) = \frac{1}{(s - 2)(s + 3)} - \frac{y(0)}{s + 3}.$$
After doing partial fractions and taking the inverse laplace transform, I get
$$y(t) = \frac{1}{5} e^{2t} u(t) - \frac{1}{5} e^{-2t} u(t) - e^{-3t} u(t) \cdot y(0)$$
Did I do this correctly?
Well, we are trying to solve:
$$ \begin{cases} \text{y}'\left(t\right)+3\cdot\text{y}\left(t\right)=\text{x}\left(t\right)\\ \\ \text{x}\left(t\right)=\exp\left(-2t\right)\cdot\theta\left(-t\right) \end{cases}\tag1 $$
Where $\theta\left(t\right)$ is the Heaviside step function.
Taking the Laplace transform of both sides, gives:
$$ \begin{cases} \text{s}\cdot\text{Y}\left(\text{s}\right)-\text{y}\left(0\right)+3\cdot\text{Y}\left(\text{s}\right)=\text{X}\left(\text{s}\right)\\ \\ \text{X}\left(\text{s}\right)=0 \end{cases}\tag2 $$
Solving for $$, gives:
$$\text{Y}\left(\text{s}\right)=\frac{\text{y}\left(0\right)}{\text{s}+3}\tag3$$