# use the Laplace transform to solve the initial value problem. y"-3y'+2y=begin{cases}0&0leq t<11&1leq t<2 -1&tgeq2end{cases} y(0)=-3 y'(0)=1

Question
Laplace transform
use the Laplace transform to solve the initial value problem.
$$y"-3y'+2y=\begin{cases}0&0\leq t<1\\1&1\leq t<2\\ -1&t\geq2\end{cases}$$</span>
$$y(0)=-3$$
$$y'(0)=1$$

2021-03-08
Step 1
It is required to solve the initial value problem using laplace transform:
$$y"-3y'+2y=\begin{cases}0&0\leq t<1\\1&1\leq t<2\\ -1&t\geq2\end{cases}$$</span>
$$y(0)=-3$$
$$y'(0)=1$$
Step 2
The first step in using laplace transform to solve an initial value problem is to take the transform of every term of differential equation:
$$L(y")-3L(y')+2L(y)=0 , 0\leq t<1$$</span>
Step 3
Now, use the appropriate formula’s:
$$s^2Y-sy(0)-y'(0)-3(sY(s)-y(0))+2(Y(s))=L(0) , 0\leq t<1$$</span>
$$s^2Y(s)-s(-3)-(1)-3sY(s)+3(-3)+2Y(s)=0$$
$$(s^2-3s+2)Y(s)=10-3s$$
$$Y(s)=\frac{10-3s}{(s^2-3s+2)} \Rightarrow Y(s)=\frac{10-3s}{(s-1)(s-2)}$$
Step 4
Now, find the partial fraction decomposition for the transform:
$$\frac{10-3s}{(s-1)(s-2)}=\frac{A}{(s-1)}+\frac{B}{(s-2)}$$
$$=\frac{A(s-2)+B(s-1)}{(s-1)(s-2)}$$
$$10-3s=As-2A+Bs-B$$
$$10-3s=(A+B)s+(-2A-B)$$
$$A+B=-3$$
$$-2A-B=10$$
after solving : A=-7 and B=4
Step 5
Thus,
$$Y(s)=\frac{10-3s}{(s-1)(s-2)}=\frac{A}{(s-1)}+\frac{B}{(s-2)}=\frac{-7}{(s-1)}+\frac{4}{(s-2)}$$
finally taking the inverse transform:
$$y(t)=-7e^t+4e^{2t}$$
Step 6
Now, consider the case when $$f(t) =1, 1\leq t<2$$</span>
$$s^2Y(s)-sy(0)-y'(0)-3(sY(s)-y(0))+2(Y(s))=L(1) , 1\leq t<2$$</span>
$$s^2Y(s)-s(-3)-(1)-3sY(s)+3(-3)+2Y(s)=\frac{1}{s}$$
$$(s^2-3s+2)Y(s)=10-3s+\frac{1}{s}$$
$$Y(s)=\frac{10-3s+\frac{1}{s}}{s^2-3s+2} \Rightarrow Y(s)=\frac{-3s^2+10s+1}{s(s-1)(s-2)}$$
Step 7
Now, find the partial fraction decomposition for the transform:
$$\frac{-3s^2+10s+1}{s(s-1)(s-2)}=\frac{A}{s}+\frac{B}{s-1}+\frac{C}{s-2}$$
$$=\frac{A(s-1)(s-2)+Bs(s-2)+C(s)(s-1)}{s(s-1)(s-2)}$$
$$-3s^2+10s+1=(A+B+C)s^2+(-3A+2B-C)s+2A$$
$$A+B+C=-3$$
$$-3A+2B-C=10$$
$$2A=1 \Rightarrow A=\frac{1}{2}$$
after solving $$B=\frac{8}{3} \text{ and } C=\frac{-45}{6}$$
Step 8
Thus,
$$Y(s)=\frac{-3s^2+10s+1}{s(s-1)(s-2)}=\frac{A}{s}+\frac{B}{s-1}+\frac{C}{s-2}=\frac{\frac{1}{2}}{s}+\frac{\frac{8}{3}}{s-1}+\frac{\frac{-45}{6}}{s-2}$$
finally taking the inverse transform
$$y(t)=\frac{1}{2}+\frac{8}{3}e^t+(\frac{-45}{6})e^{2t}$$
Step 9
Now, consider the case when $$f(t) = 1,t \Rightarrow 2$$
$$s^2Y-sy(0)-y'(0)-3(sY(s)-y(0))+2(Y(s))=L(2) t\Rightarrow 2$$
$$s^2Y(s)-s(-3)-(1)-3sY(s)+3(-3)+2Y(s)=\frac{2}{s}$$
$$(s^2-3s+2)Y(s)=10-3s+\frac{2}{s}$$
$$Y(s)=\frac{10-3s+\frac{1}{s}}{s^2-3s+2} \Rightarrow Y(s)=\frac{-3s^2+10s+2}{s(s-1)(s-2)}$$
Step 10
Now, find the partial fraction decomposition for the transform:
$$\(\frac{-3s^2+10s+2}{s(s-1)(s-2)}=\frac{A}{s}+\frac{B}{s-1}+\frac{C}{s-2}$$
$$=\frac{A(s-1)(s-2)+Bs(s-2)+C(s)(s-1)}{s(s-1)(s-2)}$$
$$-3s^2+10s+2=(A+B+C)s^2+(-3A+2B-C)s+2A$$
$$A+B+C=-3$$
$$-3A+2B-C=10$$
$$2A=2 \Rightarrow A=1$$
$$\text{after solving } B=3 \text{ and } C=-7$$
$$\text{ thus }$$
$$Y(s)=\frac{-3s^2+10s+2}{s(s-1)(s-2)}=\frac{A}{s}+\frac{B}{s-1}+\frac{C}{s-2}=\frac{1}{s}+\frac{3}{s-1}+\frac{-7}{s-2}$$
$$\text{ finally taking the inverse transform }$$
$$y(t)=1+3e^t+{-7}e^{2t}$$

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