Laplace Transform method to solve a differential equation

Question

I'm trying to solve this diffrential equation using Laplace Transform method but have trouble finishing it (x(0) = 0):

$$\frac{dx}{dt}+2x=15e^{-2t}$$ $$L\left(\frac{dx}{dt}+2x=15e^{-2t}\right)$$ $$Xs+x(0)+2X=\frac{15}{s+2}$$ $$Xs+2X=\frac{15}{s+2}$$ $$X(s+2)=\frac{15}{s+2}$$ $$X=\frac{15s+30}{s+2}$$ $$L^{-1}X=L^{-1}\frac{15s+30}{s+2}$$ $$x=L^{-1}\frac{15s}{s+2}+L^{-1}\frac{30}{s+2}$$ $$x=15L^{-1}\frac{s}{s+2}+30L^{-1}\frac{1}{s+2}$$ $$x=15L^{-1}\frac{s}{s+2}+30e^{-2t}$$

How do I find the inverse laplace of $\frac{s}{s+2}$?

Answer 1

1. You have a sign error: $L(x')=sX-x(0)$. It doesn't matter here, since $x(0)=0$.

2. You multiplied both sides by $s+2$ when you should have divided both sides by $s+2$. Thus you should have arrived at

$$X=\frac{15}{(s+2)^2}.$$

This should be easily found in your LT table.

A good rule of thumb: the Laplace transform of a "reasonable" function goes to zero as $s$ tends to infinity. So if you get something like $\frac{s}{s+2}$ that doesn't do that, you probably made a mistake somewhere.

Answer 2

You've made a couple mistakes but @Ian said that!

So my help goes with the inverse Laplacetransform:

$$\mathcal{L}_{s}^{-1}\left[\frac{s}{s+2}\right]_{(t)}=\mathcal{L}_{s}^{-1}\left[1-\frac{2}{s+2}\right]_{(t)}=\mathcal{L}_{s}^{-1}\left[1\right]_{(t)}-2\mathcal{L}_{s}^{-1}\left[\frac{1}{s+2}\right]_{(t)}=\delta(t)-2e^{-2t}$$