Solving $y'' - y = 0$ I am attemtping to solve $y'' - y = 0$
I come to this solution, by using something like
$\frac{dy}{dx} = p$
So it does
$\frac{dp}{dy} \cdot \frac{dy}{dx} - y = 0$
Which gives
$\frac{dp}{dy} \cdot p = y$
After all the transformations, integrating and all, I end up with this expression!
$c_{1}e^{x} = y + \sqrt{y^{2} + c_{2}}$
wow... How am I supposed, from there, to obtain the expected solution, that is,
$y(x) = c_{1}e^{x}+c_{2}e^{-x}$
(Note c1 and c2 are unrelated to the other equation)
Help me please... really.
Thank you
 A: Starting from $c_1e^x= y+\sqrt{y^2+c_2}$:
$$c_1e^x -y= \sqrt{y^2+c_2}$$
Taking square of both sides
$$c_1e^{2x} + y^2 - 2c_1e^xy = y^2 + c_2$$
Finally it comes to $$y = \frac{c_1^2e^{2x} + c_2}{2c_1e^x}$$
And there is your solution.
A: General hint:
Assume your solution $y(x)$ has a general form like $e^{mx}$ for some $m$. So, set $y=e^{mx}$ and then satisfy it into the ODE to find the probable value of $m$. Note that the superposition principle will help us to find the general solution for this ODE.
A: One possible way would be , let 
$$\displaystyle \frac{d}{dx}y(x) + y(x) = g(x) \hspace{1cm} (1) \\  \displaystyle \frac{d}{dx}g(x) - g(x) = 0 \hspace{1cm} (2) $$
first solve $(2)$ for $g(x)$ and then put it's value on $(1)$ and solve $y(x)$ in first order equation.
Equation $(2)$ gives you $\displaystyle g(x) = c_1 e^{x}$ and Equation $(2)$ gives you $\displaystyle y(x) = c_1 e^x + c_2 e^{-x}$. It also gives you an idea that solution will be of form $y(x)=e^{\lambda x}$.
The idea is to represent the equation in terms of operator. i.e. $(D^2 -1) y(x) = 0$. Then $(D^2 - 1)y = (D-1)(D+1)y$, let $(D+1)y = g(x)$ then you have $(D-1)g(x) = 0$ and solve like above.
A: $\newcommand{\+}{^{\dagger}}%
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With your method you'll have: $y' = p$ and $p' = y$ which is equivalent to
$$
{y' \choose p'} = \overbrace{\pars{\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}}}^{\ds{A}}{y \choose p}\quad\imp\quad{y \choose p} = \expo{Ax}{y_{0} \choose p_{0}}
$$
$\ds{A^{2} = \overbrace{\pars{\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}}}^{\ds{I}}.\quad}$ Also $\ds{\pars{\expo{Ax}}' = A\expo{Ax}}$
$\ds{\pars{\expo{Ax}}'' = \expo{Ax}}$, we'll have:
$$
\expo{Ax} = \cosh\pars{x}I + \sinh\pars{x}A 
=
\pars{%
\begin{array}{cc}
\cosh\pars{x} & \sinh\pars{x}
\\
\sinh\pars{x} & \cosh\pars{x}
\end{array}}
$$
$$\color{#0000ff}{\large%
y = y_{0}\cosh\pars{x} + p_{0}\sinh\pars{x}
=
\overbrace{\half\pars{y_{0} + p_{0}}}^{\ds{\equiv\ c_{1}}}\expo{x}
+
\overbrace{\half\pars{y_{0} - p_{0}}}^{\ds{\equiv\ c_{2}}}\expo{-x}}
$$
A: Your differential equation is a special case of the more general ode $$ay^{''}+by^{'}+cy=0,$$ where $a,b,c$ are real numbers. One way to solve it is by the following two steps:

*

*solve $at^2+bt+c=0$ in $t$;

*check if the roots are either: a) real and different, b) real and repeated or c) complex.

Case a) Let the real and different roots be $t_1,t_2$, the solution is
$$
y(x) = c_1e^{t_1x} + c_2e^{t_2x}.
$$
Case b) Let the real repeated root be $t$, the solution is
$$
y(x) = c_1e^{tx} + c_2xe^{tx}.
$$
Case c) Let the complex roots be ${t_{1,2}} = \lambda  \pm \mu \,i $, the solution is
$$
y\left( x \right) = {c_1}{{e}^{\lambda x}}\cos \left( {\mu \,x} \right) + {c_2}{{e}^{\lambda x}}\sin \left( {\mu \,x} \right).
$$
