Using Laplace Transform , solve the following differential equation {y}text{}-{4}{y}={e}^{{-{3}{t}}},{y}{left({0}right)}={0},{y}'{left({0}right)}={2}

Tahmid Knox 2020-12-01 Answered
Using Laplace Transform , solve the following differential equation
y4y=e3t,y(0)=0,y(0)=2
a) 1420e2t530e2t930e6t
b) 1120e2t5120e2t420e3t
c) 1415e2t510e2t920e3t
d) 1420e2t+520e2t920e3t
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Expert Answer

broliY
Answered 2020-12-02 Author has 97 answers
Step 1
As per guidelines, only typed answers are acceptable. so we will give only typed solution.
Given initial value problem is
y4y=e3t,y(0)=0,y(0)=2
Step 2
Applying Laplace transform on both sides
L{y4y}=L{e3t}
L{y}4L{y}=1s+3
s2L{y}sy(0)y(0)4L{y}=1s+3
(s24)L{y}2=1s+3
L{y}=1(s+3)(s24)+2s24
=(15(s+3)+120(s2)14(s+2))+(12(s+2)+12(s2))
=15(s+3)+1120(s2)34(s+2)
Taking inverse Laplace transform on both sides, we get
y=L1{15(s+3)+1120(s2)34(s+2)}
=L1{15(s+3)}+L1{1120(s2)}L1{34(s+2)}
=15e3t+1120e2t34e2t
Hence, solution is
y(t)=1120e2t34e2t+15e3t  or  y(t)=1120e2t1520e2t+420e3t
Comparing the answer with options, none of the option is correct.
The option given are all wrong.
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