Solving $y'' - y = 0$

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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:

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

$$ y\left( x \right) = {c_1}{{e}^{\lambda x}}\cos \left( {\mu \,x} \right) + {c_2}{{e}^{\lambda x}}\sin \left( {\mu \,x} \right). $$

Answer 5

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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: $$ \expo{Ax} = \cosh\pars{x}I + \sinh\pars{x}A = \pars{% \begin{array}{cc} \cosh\pars{x} & \sinh\pars{x} \\ \sinh\pars{x} & \cosh\pars{x} \end{array}} $$ $$\color{#0000ff}{\large% y = y_{0}\cosh\pars{x} + p_{0}\sinh\pars{x} = \overbrace{\half\pars{y_{0} + p_{0}}}^{\ds{\equiv\ c_{1}}}\expo{x} + \overbrace{\half\pars{y_{0} - p_{0}}}^{\ds{\equiv\ c_{2}}}\expo{-x}} $$