Question

Solve the following differential equation by the method of Laplace transform: \[ y''' + 2y'' - y' - 2y = 0 \] given that $y(0) = 0$, $y'(0) = 0$, and $y''(0) = 6$.
Choose the correct answer from the options below:

• A. $y(t) = e^t - 3e^{-t} + 2e^{-2t}$
• B. $y(t) = e^t + e^{-t} + 2e^{-2t}$
• C. $y(t) = 2e^t + 3e^{-t} - 2e^{-2t}$
• D. $y(t) = 2e^t + \sin t - \cos 3t$

Hint:

For higher-order differential equations, Laplace transforms simplify solving by converting the equation into algebraic form.

Answer 1

The Correct Option is A

Using the Laplace transform method, we take the transforms of the given differential equation and solve for $Y(s)$. After applying the initial conditions, we find:
\[y(t) = e^t - 3e^{-t} + 2e^{-2t}.\]