Question
Solve the given initial-value problem.$$y^{\prime \prime}+y=5 t e^{-3 t}, \quad y(0)=2, \quad y^{\prime}(0)=0$$.
Step 1
The Laplace transform of $y''$ is $s^2Y(s) - sy(0) - y'(0)$ and the Laplace transform of $y$ is $Y(s)$. The Laplace transform of $5te^{-3t}$ is $\frac{5}{(s+3)^2}$. So, we get: $$s^2Y(s) - sy(0) - y'(0) + Y(s) = \frac{5}{(s+3)^2}$$ Show more…
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