solve the initial value problem by the Laplace Transformation Method: y(0) = 1

 

dy + 3y = e-t   

dt                                      

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Apply Laplace Transform to each term:

L{y'(t)}+3L{y(t)}=L{e⁻ᵗ}, so sY(s)-y(0)+3Y(s)=1/(s+1).

Y(s)(s+3)=1+1/(s+1) because y(0)=1 as the initial condition.

Y(s)=(s+2)/((s+1)(s+3)).

(s+2)/((s+1)(s+3))≡A/(s+1)+B/(s+3), so s+2=As+3A+Bs+B.

Therefore, A+B=1, 3A+B=2; 3A+B-(A+B)=2-1=1, 2A=1, A=½=B.

Y(s)=½(1/(s+1))+½(1/(s+3)).

Now we apply inverse Laplace:

y(t)=½e⁻ᵗ+½e⁻³ᵗ.

by Top Rated User (694k points)

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