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
asked 2021-03-07
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>
\(y(0)=-3\)
\(y'(0)=1\)

Answers (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:
\( \(\frac{-3s^2+10s+2}{s(s-1)(s-2)}=\frac{A}{s}+\frac{B}{s-1}+\frac{C}{s-2}\)
\(=\frac{A(s-1)(s-2)+Bs(s-2)+C(s)(s-1)}{s(s-1)(s-2)}\)
\(-3s^2+10s+2=(A+B+C)s^2+(-3A+2B-C)s+2A\)
\(A+B+C=-3\)
\(-3A+2B-C=10\)
\(2A=2 \Rightarrow A=1\)
\(\text{after solving } B=3 \text{ and } C=-7\)
\(\text{ thus }\)
\(Y(s)=\frac{-3s^2+10s+2}{s(s-1)(s-2)}=\frac{A}{s}+\frac{B}{s-1}+\frac{C}{s-2}=\frac{1}{s}+\frac{3}{s-1}+\frac{-7}{s-2}\)
\(\text{ finally taking the inverse transform }\)
\(y(t)=1+3e^t+{-7}e^{2t}\)
0

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