Question

Let x(t) be the solution of the initial-value problem
(a) Find the Laplace transform F(s) of the forcing f(t).
(b) Find the Laplace transform X(s) of the solution x(t).
\(x"+8x'+20x=f(t)\)
\(x(0)=-3\)
\(x'(0)=5\)
\(\text{where the forcing } f(t) \text{ is given by }\)
\(f(t) = \begin{cases} t^2 & \quad \text{for } 0\leq t<2 ,\\ 4e^{2-t} & \quad \text{for } 2\leq t < \infty . \end{cases}\)

Answer 1

given differential equation is x
\(x"+8x'+20x=f(t)\)
\(x(0)=-3\)
\(x'(0)=5\)
\(\text{where the forcing } f(t) \text{ is given by }\)
\(f(t) = \begin{cases} t^2 & \quad \text{for } 0\leq t<2 ,\\ 4e^{2-t} & \quad \text{for } 2\leq t < \infty . \end{cases}\)</span>
\(\therefore \ x"+8x'+20x=t^2+4e^{2-t}\)
\(\text{we know that } L\left[x"\right]=s^2X(s)-sX(0)-x'(0)\)
\(L\left[x'\right]=sX(s)-x(0)\)
\(L\left[t^n\right]=\frac{n!}{s^{n+1}} , L\left\{e^{-at}\right\}=\frac{1}{s+a}\)
\(\text{Step 2}\)
\(\text{Taking Laplace tranform of equation}\)
\(L\left[x"\right]+8L\left[x'\right]+20L\left[x\right]=L\left[t^2\right]+4e^2L\left\{e^{-t}\right\}\)
\(s^2(X(s))-sX(0)-x'(0)+8\left[s(X(s))-x(0)\right]+20X(s)=\frac{2!}{s^{2+1}}+\frac{4e^{2}}{s+1}\)
\(\therefore s^2X(s)-s(-3)-5+8(sX(s)-5)+20X(s)=\frac{2}{s^{3}}+\frac{4e^{2}}{s+1}\)
\(\therefore (s^2+8s+20)X(s)+3s-5-40=\frac{2}{s^{3}}+\frac{4e^{2}}{s+1}\)
\(\therefore (s^2+8s+20)X(s)+3s-45=\frac{2}{s^{3}}+\frac{4e^{2}}{s+1}\)
\(\therefore X(s)\left[s^2+8s+20\right]=\frac{2}{s^{3}}+\frac{4e^{2}}{s+1}+45-3s\)
\(\therefore x(s)=\frac{1}{s^2+8s+20}\left\{(45-3s)+\frac{2}{s^{3}}+\frac{4e^{2}}{s+1}\right\}\)