Solving differential equations by Laplace Transform

Question

. In solving the initial value problem:

dx/dt + x = f(t), x(0) = 5 where f(t) (0, 0 ≤ t < π, 3cost, t ≥ π,) find the Laplace transform Xˆ(s) and hence x(t)

Answer 1

HINT, to solve:

$$x'(t)+x(t)=f(t)$$

Use Laplace transform:

$$\text{s}\text{X}(\text{s})-x(0)+\text{X}(\text{s})=\text{F}(\text{s})$$

Now, we know that $x(0)=5$:

$$\text{s}\text{X}(\text{s})-5+\text{X}(\text{s})=\text{F}(\text{s})$$

Solving for $\text{X}(\text{s})$:

$$\text{X}(\text{s})=\frac{\text{F}(\text{s})+5}{1+\text{s}}$$

With invere Laplace transform:

$$x(t)=\mathcal{L}_\text{s}^{-1}\left[\frac{\text{F}(\text{s})}{1+\text{s}}\right]_{(\text{s})}+5e^{-t}$$

Answer 2

Start off by expressing $f(t)$ in terms of the unit step function $u_c(t)$. You have the following:

$$f(t) = 0 , 0≤t<π$$ $$f(t) = 3cos(t) , t≥π$$

This is what the function looks like (http://m.wolframalpha.com/input/?i=θ%28t-π%29%283cost%29&x=0&y=0).

I pretty much already gave you the answer in wolfram, you express $f(t)$ in terms of the unit step function as follows:

$$f(t) = 3u_π(t)cos(t)$$

$u_π(t)$ is equal to zero as long as $0≤t<π$, and it is equal to one as long as $t≥π$. This satisfies the conditions for $f(t)$ if you wanna check.

Now the equation becomes:

$$x'(t)+x(t)=3u_π(t)cos(t)$$

Take the Laplace Transform on both sides:

$$\mathcal{L}(x'(t))+\mathcal{L}(x(t))=\mathcal{L}(3u_π(t)cos(t))$$

Notice that $cos(t) = -cos(t-π)$ and thus:

$$sF(s) - x(0) + F(s) = -3\mathcal{L}(u_π(t)cos(t-π))$$ $$sF(s) + F(s) - 5 = -3e^{-πs}\frac{s}{s^2+1}$$ $$F(s)(s+1) =\frac{-3e^{-πs}s}{s^2+1}+5$$ $$F(s)=\frac{-3e^{-πs}s}{(s+1)(s^2+1)}+\frac{5}{s+1}$$ $$F(s)=\frac{-3e^{-πs}s+5(s^2+1)}{(s+1)(s^2+1)}$$

Now you have to take the inverse Laplace Transform of $F(s)$ to get

$$\mathcal{L^{-1}}(F(s))=x(t)$$

I'll leave this one for you to do!