# 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

• Hint: Does the auxiliary polynomial, or characteristic polynomial ring a bell? y''-y=0 gives us the polynomial r^2-1=0 – EpicMochi Nov 6 '13 at 9:43
• do you just want the solution (in some other manner) or you want to use just this method... :O – user87543 Nov 6 '13 at 9:44
• Thank you for your comments but... I am supposed to get it by this solution. I mean... the solutions book arrived to the same thing than I. They just... come from the solution I have (well, some constants differ but its essentially the same), but they do not show how they went from the euqation I gave to the final solution :( So basically, yes, I'm supposed to solve it this way – Yannick Nov 6 '13 at 9:46

## 4 Answers

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.

• Sigh. I had it in my face all this time. I dunno why, but from first glace I told myself "you cannot get rid of y^2". Guess I am tired. Thank you – Yannick Nov 6 '13 at 10:27

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.

• Yes, I have seen this mentionned multiples times before. Well, on different sites and all. It brings an interesting subject actually and a question I had about it : on which cases can we apply this assumption? On equations of all forms or... only this particular form?! – Yannick Nov 6 '13 at 10:01
• @Yannick: You can use this way for a linear $n$th-order OE with constant coefficients. – Mikasa Nov 6 '13 at 10:08
• Ohhhh I actually get it now. I shud have read a beeet moar before trying to solve a problem :) Thanks mate – Yannick Nov 6 '13 at 14:01
• @Yannick: Thanks. However, other posts completed the answer. :-) – Mikasa Nov 6 '13 at 14:04
• $\ddot\smile \rightarrow +1$ – amWhy Nov 6 '13 at 14:22

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.

• I did not think about this one. Very nice. I was actually sticking with solutions similar with what I've read so far, but I guess we can use any substitution. Thanks for this one. – Yannick Nov 6 '13 at 10:24
• @Yannick you are welcome :) – Santosh Linkha Nov 6 '13 at 10:27

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

• That's pretty fantastic, I'd never thought of exponentiating matrices as a method of solving differential equations! – user18862 Jan 9 '14 at 4:47
• @NeuroFuzzy This is a good way to have first order equations. The price to pay is "matrices". Thanks. – Felix Marin Jan 9 '14 at 4:49