use the Laplace transform to solve the given initial-value problem. y"+y'-2y=10e^{-t}, y(0)=0,y'(0)=1

use the Laplace transform to solve the given initial-value problem. y"+y'-2y=10e^{-t}, y(0)=0,y'(0)=1

Question
Laplace transform
asked 2020-12-17
use the Laplace transform to solve the given initial-value problem. \(y"+y'-2y=10e^{-t}, y(0)=0,y'(0)=1\)

Answers (1)

2020-12-18
Step 1
Given \(y"+y'-2y=10e^{-t} - (1)\)
and \(y(0)=0,y'(0)=1\)
By taking laplace transform of (1)
\(S^2y(s)-Sy(0)-y'(0)+sy(s)-y(0)-2y(s)=\frac{10}{S+1}\)
Step 2
\(\Rightarrow (S^2+S-2)y(S).-S.0-1-0=\frac{10}{s+1}\)
\(\Rightarrow (S^2+S-2)y(S)=1+\frac{10}{s+1}\)
\(\rightarrow Y(s)=\frac{1}{S^2+S-2}+\frac{10}{(S+1)(S^2+S-2)}\)
\(=\frac{1}{(S+2)(S-1)}+\frac{10}{(S+1)(S-1)(S+2)}\)
\(=-\frac{1}{3(S+2)}+\frac{1}{3(S-1)}+\frac{5}{3(S-1)}-\frac{5}{S+1}+\frac{10}{3(S+2)}\)
\(Y(s)=\frac{3}{S+2}+\frac{2}{S-1}-\frac{5}{S+1}\)
\(\therefore y(t)=e^{-2t}+e^t-e^{-t}\) for t>0
0

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