Solving Differential Equation Multiple Ways I am currently self learning differential equations and I use the book Elementary Differential Equations. My question is that I saw many ways to solve a DE. Can I use any method to solve any DE?
For example, if I had the DE: $y''+5y'-6y=10e^x$ and suppose it is an IVP so lets say $y(0)=1$ and $y'(0)=1$. Can someone show me the solution to this using the annihilation method, variation of parameters method, and Laplace method?
 A: Laplace Transform Method
Given the ODE $$y''+5y'-6y=10e^x$$The initial conditions which are the following : $y(0)=1 \ \ \ \ y'(0)=1$
Then I perform the indicated Laplace Transform $\mathcal{L}(y(x))$:
\begin{align}\mathcal{L}(y''+5y'-6y)&=\mathcal{L}(10e^x) \\s^2Y(s)-s*y(0)-y'(0)+5(sY(s)-y(0))-6Y(s)&=\frac{10}{s-1} \\ (s^2+5s-6)Y(s)&=s+6+\frac{10}{s-1}\\ Y(s)&=\frac{1}{s-1}+\frac{10}{(s-1)^2(s+6)}\\Y(s)&=\frac{1}{s-1}-\frac{10}{49}\frac{1}{s-1}+\frac{10}{49}\frac{1}{s+6}+\frac{10}7\frac{1}{(s-1)^2} \end{align}
From here then I take the inverse Laplace transform to get following:
\begin{align}y&=\frac{39}{49}e^x+\frac{10}{49}e^{-6x}+\frac{10}7xe^x\end{align}
Wolfram Alpha got the same answer and Desmos graphs says both solutions are the same
A: Undetermined Coefficients-Superposition Approach
To first attempt this ODE you'll have to solve the associated homogeneous equation which is the following. $$y''+5y'-6y=0$$
To begin to solve that one does the following by writing down the auxiliary equation:
\begin{align}m^2+5m-6&=0\\ (m+6)(m-1)&=0\\m&=-6,1\end{align}
Using these two solutions one can get the following homogenous solution:
$$y_h=c_1e^{-6x}+c_2e^x$$
The only thing that is left is solve for the particular of the solution which is the following:
$$y_p=Axe^x$$
This was done because if we used $y_p=Ae^x$, it would not work.
\begin{align}y_p''+5y_p'-6y_p&=10e^x\\ 2Ae^x+Axe^x+5Ae^x+5Axe^x-6Axe^x&=10e^x\\ 7Ae^x&=10e^x\\ A&=\frac{10}7\end{align}
This means that our particular solution is $y_p=\frac{10}7xe^x$
Then if you add the homogeneous with the particular:
\begin{align}y&=y_h+y_p \\ y&=c_1e^{-6x}+c_2e^x+\frac{10}7xe^x\end{align}
Then using the inital conditions and setting up our equations we get the following system of equations to find the coefficients.
\begin{align}1&=c_1+c_2 \\ 1&= -6c_1+c_2+\frac{10}7\end{align}
Once we solve those systems we get that the DE with initial conditions is:
$$y=\frac{39}{49}e^x+\frac{10}{49}e^{-6x}+\frac{10}{7}xe^x$$
A: Annihilator Approach
$$y''+5y'-6y=10e^x$$
To begin the problem we will make the substitution of $D$, and use $D-1$ to annihilate $10e^x$ So one would proceed as follows:
\begin{align}(D-1)(D^2+5D-6)&=10(D-1)e^x\\ D^3+4D^2-11D+6&=0\end{align}
Now using the auxiliary equation equation we can setup the problem as follows, and based on other posts its very similar setup.
\begin{align}m^3+4m^2-11m+6&=0\\ (m-1)^2(m+6)&=0\end{align}
I achieved that factorization using Possible Rational Zeroes:
$$\text{PRZs}=\pm1,\pm2,\pm3,\pm6$$
The solution $m=1$ works so then synthetic divide:
\begin{array}{c|cccc}1&1&4&-11&6\\ & & 1 & 5 &-6\\ & \hline  1 &5&-6&0\end{array}
Then the complete factored form is the following $(m-1)^2(m+6)$. The solutions then being the following $m=1,1,-6$ as we know for repeated factors one goes up in linearity.
$$y=c_1e^x+c_2xe^x+c_3e^{-6x}$$
Annihilator approach is little bit counter intutitve in my eyes, because you are kind of forced to identify the homogeneous component of the answer. Of course we solved for that homogeneous component already which is $y_h=c_1e^x+c_3e^{-6x}$ Then we solve for the particular the same way as the other method the Superposition approach for $y_p$, and then using the intial conditions we get the same answers:
\begin{align}y&=\frac{39}{49}e^x+\frac{10}{49}e^{-6x}+\frac{10}7xe^x\end{align}
