Solve $y''-y'-6y=e^{3t}+5$ 
Solve the following initial value problem $$y'' - y' - 6y = e^{3t} + 5, \quad y(0) = 0, \quad y'(0) = 0 $$


I used the Laplace transform and got
$$ Y(s) \left(s^2-s-6\right)=\dfrac 1{s-3}+\dfrac 5s $$
then brought it to the other side
$$Y(s)=\dfrac 1{(s-3)(s^2-s-6)}+\dfrac 5{s(s^2-s-6)}$$
Taking user577215664's advice, I got
$$s^2−s−6=(s+2)(s−3)$$
so i split it into partial fractions
$$\dfrac A{(s-3)}+\dfrac B{(s-3)^2}+\dfrac C{(s+2)}$$ and solving it i got
in part A
$$ A=-\dfrac1{17}, \qquad B = C = \dfrac1{17} $$
and in part b
i got
$$ B=1\dfrac4{6}, \qquad A = C = -\dfrac5{6} $$
so end up having
$$-\dfrac1{17(s-3)}+\dfrac1{17(s-3)^2}+\dfrac1{17(s+2)}+-\dfrac5{6(s)}+\dfrac{10}{6(s-3)}+-\dfrac5{6(s+2)}$$
$$\dfrac1{17}e^{3t}+\dfrac1{17}te^{3t}+\dfrac1{17}e^{-2t}+-\dfrac5{6}+\dfrac{10}{6}e^{3t}+-\dfrac5{6}e^{-2t}$$
which can be simplified to
$$\dfrac{82}{51}e^{3t}+\dfrac{79}{102}e^{-2t}+\dfrac1{17}te^{3t}+\dfrac5{6}$$
is this correct?
 A: Let's go to try to follow in your steps :

*

*We goal is to find the solution for the problem $y''-y'-6y=e^{3t}+5$ with $y(0)=y'(0)=0$.

*Laplace transform $L$ over the differential equation return a algebraic equation
$$L(y''-y'-6y)=L(e^{3t}+5)$$
by linearity and using the fact $y(0)=y'(0)=0$.
Then,
$$(S^{2}-S-6)L(y(t))=\frac{5}{S}+\frac{1}{S-3}$$

*Then, partial fraction descomposition give
$$L(y(t))=\frac{3(2S-5)}{S(S-3)^{2}(S+2)}=\frac{1/5}{(S-3)^{2}}+\frac{22/75}{(S-3)}+\frac{-5/6}{S}+\frac{27/50}{(S+2)}$$

*Using the fact $L^{-1}(L(y(t))(s)=y(t)$, then
$$y(t)=L^{-1}\left(\frac{1/5}{(S-3)^{2}}+\frac{22/75}{(S-3)}+\frac{-5/6}{S}+\frac{27/50}{(S+2)} \right),$$
againg by linearity and using the known Laplace transforms, then we come to the particular solution for the problem given
$$y(t)=\frac{22}{75}e^{3t}+\frac{1}{5}te^{3t}+\frac{27}{50}e^{-2t}-\frac{5}{6}.$$
Note :

*

*Your solution is not correct but fortunately, the problem does not seem to be in theory, perhaps is just some miscalculation, for example we can check that your solution does not work since if $t=0$ in your function $y(t)=\dfrac{82}{51}e^{3t}+\dfrac{79}{102}e^{-2t}+\dfrac1{17}te^{3t}+\dfrac5{6}$ you have $y(0)\not=0$ and it is a contradiction with the initial condition $y(0)=0$.

PS : For the partial fraction descomposition, consider that $\frac{3(2S-5)}{S(S-3)^{2}(S+2)}$ can be written as
$$\frac{3(2S-5)}{S(S-3)^{2}(S+2)}=\frac{A}{S-3}+\frac{B}{(S-3)^{2}}+\frac{C}{S}+\frac{D}{S+2}$$
Then, we can work with a bit of algebra and we get the linear system of equations
$$\begin{align} 
1\to\\S\to\\S^{2}\to\\S^{3}\to\\\end{align}\begin{pmatrix}0&0&18&0\\-6&2&-3&9\\-1&1&-4&-6\\1&0&1&1\end{pmatrix}\begin{pmatrix} A\\ B\\ C\\ D\end{pmatrix}=\begin{pmatrix}-15\\6\\0\\0\end{pmatrix}$$
Solving the linear system of equations we have the solution
$$(A,B,C,D)=(22/75,1/5,-5/6,27/50)$$
