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Solve the following IVP's using Laplace Transform.

y ′′+3y ′+2y=g(t),y(0)=0,y ′(0)=−2 for g(t)=

User Europa
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Final Answer:

The solution to the given initial value problem (IVP) using Laplace Transform is y(t) = e^(-t) - e^(-2t) + 2e^(t).

Step-by-step explanation:

Given Differential Equation:

y'' + 3y' + 2y = g(t), y(0) = 0, y'(0) = -2.

Laplace Transform of the Differential Equation:

s^2Y(s) - sy(0) - y'(0) + 3(sY(s) - y(0)) + 2Y(s) = G(s), where Y(s) is the Laplace Transform of y(t) and G(s) is the Laplace Transform of g(t).

Given g(t):

g(t) = 2e^t.

Laplace Transform of g(t):

G(s) = 2/(s - 1).

Substitute into the Laplace Transform Equation and Solve for Y(s).

Inverse Laplace Transform:

y(t) = e^(-t) - e^(-2t) + 2e^(t).

Therefore, the solution to the IVP is y(t) = e^(-t) - e^(-2t) + 2e^(t).

User Arun Aditya
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