I have the following homework problem. Using Laplace transform, I need to solve the following differential equation
$$
y'' + 8y' + 15y = 35 \theta(2-x)e^{2x}
$$
where $\theta(x)$ is the Heaviside step function. I evaluated this differential equation using WolframAlpha, but the solution it presents is for the homogeneous equation, which is not the same as the equation I have since for $2-x>0$ the R.H.S. is $35 e^{2x}\neq 0$. Since I believe there's no general formula for the Laplace transform of a product, I explicitly calculated the Laplace transform of the R.H.S. which gives
$$
\mathcal{L}\left\{35 \theta(2-x)e^{2x} \right\} = 35 \int_{0}^{2}e^{-px}e^{2x} \ dx = \frac{35\left(1-e^{4-2s}\right)}{s-2} \tag{1}
$$
Which would then imply, by taking the Laplace transform on both sides of the original equation, that
$$
Y = \frac{1}{(s+3)(s+5)}\left[\frac{35\left(1-e^{4-2s}\right)}{s-2} +y(0)(s+8) +y'(0) \right]\tag{2}
$$
Now, for the last $2$ summands I know I can apply partial fractions to get the inverse Laplace transform, but for the first term, I have no idea how to inverse Laplace it. I plugged it into WA and got a really long expression for the answer. This last part seemed odd to me and I started doubting if my approach was correct.
I then had the idea that maybe I should try to solve the equation piece-wise, getting $2$ "simpler" equations for $x<2$ and $x>2$, the latter being the homogeneous case. The problem with this is that I don't think this results in a solution to the original problem since in both cases I would get different functions, and I need a function that satisfies both conditions ($x>2$ and $x<2$) simultaneously.
Am I on the right path to solving this problem? Or is my approach wrong? And in any case, is there a simple way to obtain the inverse Laplace transform of equation $(2)$? Thank you!
Best Answer
$$Y = \frac{1}{(s+3)(s+5)}\left[\frac{35\left(1-e^{4-2s}\right)}{s-2} +...... \right]$$ For the first term you can use the formula: $$\mathcal {L^{-1}}\{ e^{-cs}F(s)\}=H(t-c)f(t-c)$$ You have that: $$e^{4-2s}=e^4e^{-2s} \implies c=2$$ And: $$F(s) =- \frac{35e^4}{(s+3)(s+5)(s-2)}$$