Question

Solve by Laplace transforms y"-3y'+2y=2delta(t-1) y(0)=1 y'(0)=0

Laplace transform
ANSWERED
asked 2020-11-10
Solve by Laplace transforms
\(y"-3y'+2y=2\delta(t-1)\)
\(y(0)=1\)
\(y'(0)=0\)

Answers (1)

2020-11-11
\(\text{Step 1}\)
\(\text{The given equation is: }\)
\(y"-3y'+2y=2\delta(t-1)\)
\(\text{Applying the Laplace transform to both sides of the equation, we get }\)
\(s^2Y(s)-sy(0)-y'(0)-3sY(s)-y(0)+2Y(s)=2e^{-s}\)
\(\text{Step 2}\)
\(\text{Substituting the values given in the question, we get }\)
\(s^2Y(s)-sy(0)-y'(0)-3sY(s)+y(0)+2Y(s)=2e^{-s}\)
\(\text{or, } Y(s)\left[s^2-3s+2\right]-s+1=2e^{-s}\)
\(\text{or, }Y(s)=\frac{2e^{-s}+s-1}{s^2-3s+2}\)
\(\text{or, }Y(s)=\frac{2e^{-s}+s-1}{s^2-2s-s+2}\)
\(\text{or, }Y(s)=\frac{2e^{-s}+s-1}{(s-1)(s-2)}\)
\(\text{Step 3}\)
\(\text{Now, using partial fraction method: }\)
\(\frac{2e^{-s}+s-1}{(s-1)(s-2)}=\frac{A}{s-1}+\frac{B}{s-2}\)
\(\text{or, } \frac{2(1-s)+s-1}{(s-1)(s-2)}= \frac{A(s-2)+B(s-1)}{(s-1)(s-2)}\)
\(\text{or, } \frac{1-s}{(s-1)(s-2)}= \frac{s(A+B)-(2A+B)}{(s-1)(s-2)}\)
\(\text{Comparing the coefficients:}\)
\(A+B=-1 \text{ and } 2A+B=-1\)
\(\text{Solving the above two equations for the value of A and B, we get}\)
\(A=0 \text{ and } B=-1\)
\(\text{Step 4}\)
\(\text{Substituting the value of A and B in the equation }\frac{2e^{-s}+s-1}{(s-1)(s-2)}=\frac{A}{s-1}+\frac{B}{s-2} \text{ we get}\)
\(\frac{2e^{-s}+s-1}{(s-1)(s-2)}=\frac{-1}{s-2}\)
\(\text{taking inverse Laplace transform both sides, we get}\)
\(L^{-1}\left[\frac{2e^{-s}+s-1}{(s-1)(s-2)}\right]=L^{-1}\left[\frac{-1}{s-2}\right]\)
\(\text{or } y(t)=-e^{2t}\)
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