How can I solve this second order ODE using Laplace transform?

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!

$$Y = \frac{1}{(s+3)(s+5)}\left[\frac{35\left(1-e^{4-2s}\right)}{s-2} +...... \right]$$ 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: $$F(s) =- \frac{35e^4}{(s+3)(s+5)(s-2)}$$
Generally there is no simpler way to do the inverse Laplace transform of equation (2), and the answer provided by WA seems nothing suspicious. The Heaviside function corresponds to the exponential term in (2), and the three exponential functions of $$t$$ come from the three poles in (2).