Solve the Following Differential Equation: $y'''+y''=y'+y$ I am trying to solve the following differential equation: $y'''+y''=y'+y$
My attempt
$y'''+y''=y'+y \Leftrightarrow y'''+y''-y'-y = 0 $
I'm trying to find y in the form: $y(x) = e^{\lambda x }$
$y'''=\lambda^{3}e^{\lambda x}$
$y''=\lambda^2e^{\lambda x}$
$y' = \lambda e^{\lambda x}$
$y=e^{\lambda x}$
Therefore, $y'''+y''-y'-y = 0  \Leftrightarrow (\lambda^3+\lambda^2-\lambda-1)e^{\lambda x}$
Hence, $y(x) = e^{\lambda x}$ is a solution of  $y'''+y''-y'-y = 0 $ iff $\lambda$ is a root of the characteristic equation.
The characteristic equation $x(\lambda)= \lambda^3+\lambda^2-\lambda-1 = (x-1)(x+1)^2= 0$ Therefore $\lambda_1 = 1; \lambda_2 = -1$;
Hence $y(x) = c_1e^x+c_2e^{-x}$
How do I find $c_1$ and $c_2$ knowing that my initial conditions are $y(0)=y'(0)=0, y''(o) = 1$
edit
According to the comments: $y(x) = c_1e^x+c_2e^{-x}+c_3xe^{-x}$. Why do we add $c_3xe^{-x}$ and not simply $c_3e^{-x}$?
Also,do we have to find those constants or can we leave $c_1$ $c_2$ and $c_3$ in the final answer?
 A: Trick is: if you let $u = y'+y \implies u'' = u$. I am sure you can take it from here.
A: The characteristic polynomial is $X^3+X^2-X-1=(X-1)(X+1)^2$, so you're missing the solution $xe^{-x}$.
Consider the simpler equation $y''+2y'+y=0$. Then both $f(x)=e^{-x}$ and $g(x)=xe^{-x}$ are solutions. For the first the check is easy; for the second one you have
\begin{align}
g'(x)&=(1-x)e^{-x}\\[6px]
g''(x)&=(x-2)e^{-x}
\end{align}
and
$$
(x-2)e^{-x}+2(1-x)e^{-x}+xe^{-x}=0
$$
More generally, a factor $(X-\lambda)^k$ in the characteristic polynomial, with $k>1$, produces the linearly independent solutions
$$
e^{\lambda x},\quad
xe^{\lambda x},\quad
\dots,\quad
x^{k-1}e^{\lambda x}
$$
so $k$ linearly independent solutions.
Thus the general solution of your equation is
$$
c_1e^x+c_2e^{-x}+c_3xe^{-x}
$$
and you need initial conditions to determine $c_1$, $c_2$ and $c_3$.
With the initial condition $y(0)=0$, $y'(0)=0$ and $y''(0)=1$, you have
$$
\begin{cases}
c_1+c_2=0\\[4px]
c_1-c_2+c_3=0\\[4px]
c_1+c_2-2c_3=1
\end{cases}
$$
by just computing the derivatives and substituting $x=0$. This gives $c_1=1/4$, $c_2=-1/4$, $c_3=-1/2$.
