Solving 2nd Order ODE w/Laplace Transforms

Question

I am having difficulty with this problem:

*Note: The Delta3(t) is the delta dirac function, also the answer in the image is WRONG.

Attempt at solution :

1. Let Laplace{y(t)}=Y
2. Take Laplace of LHS and RHS.
3. Solve for Y.
4. Take inverse Laplace of Y, giving me a function of y(t)= some function

Now, are my steps correct for this type of problem?

How do I write this answer as a Heaviside function, as posed in the question?

Answer 1

Define the Laplace transform as \begin{align} f(s) = \int_{0}^{\infty} e^{-s t} \, y(t) \, dt \end{align} then for the differential equation \begin{align} y'' - 3 y' -4 y = -4 t + \delta(t), \end{align} where $y(0) = -2$ and $y'(0) = -1$ then \begin{align} s^{2} f(s) - y'(0) - s y(s) - 3s f(s) + 3 y(0) - 4 f(s) = - \frac{4}{s^{2}} + 1 \end{align} which simplifies to \begin{align} (s+1)(s-4) f(s) = 6 - 2s - \frac{4}{s^2}. \end{align} Solving for $f$ leads to \begin{align} f(s) &= \frac{6}{(s+1)(s-4)} - \frac{2 s}{(s+1)(s-4)} - \frac{4}{s^{2}(s+1)(s-4)} \\ &= \frac{6}{5} \left( \frac{1}{s-4} - \frac{1}{s+1} \right) - \frac{2}{5} \left( \frac{1}{s+1} + \frac{4}{s-4} \right) + \frac{1}{ s^2} - \frac{3}{4 s} + \frac{4}{5 (s+1)} - \frac{1}{20 (s-4)} \\ &= \frac{1}{s^{2}} - \frac{3}{4 s} - \frac{9}{20} \frac{1}{s-4} - \frac{4}{5} \frac{1}{s+1}. \end{align} Inversion provides \begin{align} y(t) = t - \frac{3}{4} - \frac{4}{5} e^{-t} - \frac{9}{20} e^{4t}. \end{align}

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The fraction involving $s^{2} (s+1)(s-4)$ is separated as follows \begin{align} \frac{1}{s^{2}(s+1)(s-4)} &= \frac{a}{s^{2}} + \frac{b}{s} + \frac{c}{s+1} + \frac{d}{s-4} \\ &= \frac{a + b s}{s^{2}} + \frac{c}{s+1} + \frac{d}{s-4}. \end{align} Now \begin{align} d &= \left.\frac{(s-4)}{s^2 (s+1)(s-4)}\right|_{s=4} = \frac{1}{80} \\ c &= \left. \frac{s+1}{s^2 (s+1)(s-4)} \right|_{s=-1} = - \frac{1}{5} \\ a &= \left. \frac{s^2}{s^2 (s+1)(s-4)} \right|_{s=0} = - \frac{1}{4} \\ b &= \frac{3}{16} \end{align} for which \begin{align} \frac{1}{s^{2}(s+1)(s-4)} = - \frac{1}{s^{2}} + \frac{3}{16 \, s} - \frac{1}{5(s+1)} + \frac{1}{80 \, (s-4)} \end{align}