The Laplace transform L {e^(-t^2)} exists, but without finding it solve the initial-value problem y''+9y=3e^(-t^2), y(0)=0,y'(0)=0

Rui Baldwin 2021-09-22 Answered
The Laplace transform L {et2} exists, but without finding it solve the initial-value problem y+9y=3et2,y(0)=0,y(0)=0
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Expert Answer

Yusuf Keller
Answered 2021-09-23 Author has 90 answers
Apply the Laplace transform to both sides of the equation:
L(y+9y)=L(3et2)
s2Y(s)sy(0)y(0)+9Y(s)=L(3et2)
(s2+9)Y(s)=L(3et2)
Y(s)=L(3et2)s2+9=13L(3et2)3s2+9
Y(s)=L(et2)xL(sin(3t))
Using the property of convolution:
f(t)g(t)LF(s)G(s)
y(t)=et2sin(3t)
y(t)=0teτ2(3(tτ))dτ
Result would be:
y(t)=0teτ2(3(tτ))dτ
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