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

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

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