Solution of $y''+5y'+6y=2e^{-x}$ I need to find the solution for $y''+5y'+6y=2e^{-x}$.
I already found the $Yh y=Ae^{-2x}+Be^{-3x}$.
But I am stuck to find the $Yp$. How can I find it?
I just assume $Y=kxe^{-x}$ is solution and find the $Y'$ and $Y''$.
But after I substitute $Y$, $Y'$, and $Y''$ to the first equation, I am stuck to $3k+2kx = 2$. What should I do?
It is said that the $Yp$ is $e^{-x}$.How?
 A: To try a particular solution (Particular Integral) of this differential equation, you should look for terms involved in $f(x)=2e^{-x}$ and its derivatives and those would be only be the constant multiples of $e^{-x}$
So, we take $y_p=ce^{-x}$, where c is a constant.
Here, it is noted that there is no duplication between the terms of $y_p$ and the terms involved in the complementary function of the given non-homogeneous equation. ($\because e^{-x}$ and either of $e^{-2x},e^{-3x}$ are linearly independent.) So, you don't have to multiply each term of $y_p$ by any power of $x$ to eliminate duplication.
After that, we substitute $y_p=ce^{-x}$ for $y$,  $y'_p=-ce^{-x}$ for $y'$, $y''_p=ce^{-x}$ for $y''$ in the non-homogeneous equation $y''+5y'+6y=2e^{-x} \tag{i}$
to get, $2ce^{-x}=2e^{-x} \implies c=1$
$\therefore y_p=e^{-x}$ satisfies the given differential equation (i).
In case you think that a term like $kxe^{-x}$ should be there in the particular solution, you can add that too but you can't drop the term $ce^{-x}$ which should be there to satisfy the differential equation (i). Thus by taking $y_p=kxe^{-x}+ce^{-x}$ and after doing the substitutions you'll see that this $y_p$ too satisfies (i) for $k=0, c=1$. The coefficient of any term which should not be there in the P.I. will eventually turn out to be zero.
The complementary function of (i) you got, $y_c=Ae^{-2x}+Be^{-3x}$ is correct. Hence the general solution of (i) is given by $y=y_c+y_p$
$\therefore y= Ae^{-2x}+Be^{-3x}+e^{-x}$ is the required general solution.
A: In general when the equation is of the form
$\phi(D)y=(D-m_{1})(D-m_{2})...(D-m_{n})y = e^{ax}$ where $D\equiv \frac{d}{dx}$ and $m_{1},m_{2},...m_{n}$ are non repeated distinct roots of the auxiliary equation and $\phi(D)$ denotes the polynomial in $D$ then the complimentary function(C.F) is given by :-
$C_{1}e^{m_{1}x}+C_{2}e^{m_{2}x}+...+C_{n}e^{m_{n}x}$ .
And the Particular Integral(P.I) = $\frac{1}{\phi(D)}e^{ax}=\frac{e^{ax}}{\phi(a)}$. Provided $\phi(a)$ is not zero. In cases of repeated roots and a being a zero of the polynomial of $kth-$order then there is a method to generalize this.
For your particular question the $\phi(D)=(D+2)(D+3)y=2e^{-x}$
So the solution is y=CF+PI :-
$$\displaystyle y=C_{1}e^{-2x}+C_{2}e^{-3x}+\frac{1}{(D+2)(D+3)}2e^{-x}=C_{1}e^{-2x}+C_{2}e^{-3x}+\frac{2}{(-1)\cdot (-2)}e^{-x}$$ .
You should read up on the operator method of solving differential equations. That would make things easier and more systematic to deal with.
A: Well, let's solve a more general DE using Laplace transform:
$$x''(t)+\text{n}x'(t)+(\text{n}+1)x(t)=\text{k}\exp\left(-x\right)\tag1$$
Taking the Laplace transform of both sides, gives:
$$\text{s}^2\text{X}\left(\text{s}\right)-\text{s}x(0)-x'(0)+\text{ns}\text{X}\left(\text{s}\right)-x(0)+(\text{n}+1)\text{X}\left(\text{s}\right)=\frac{\text{k}}{1+\text{s}}\tag2$$
Solving for $\text{X}\left(\text{s}\right)$ yields:
$$\text{X}\left(\text{s}\right)=\frac{\frac{\text{k}}{1+\text{s}}+\text{s}x(0)+x'(0)+\text{n}x(0)}{\text{s}^2+\text{ns}+\text{n}+1}=\frac{\frac{\text{k}}{1+\text{s}}+(\text{s}+\text{n})x(0)+x'(0)}{\text{s}^2+(\text{s}+1)\text{n}+1}\tag3$$
I let you find the inverse Laplace transform.
