Differential equations for this problem? How would i go about solving the differential equation> $$y''+y=0$$ (Identify the auxiliary equation.)
 A: This is called a homogeneous second-order linear differential equation. To solve, plug $y = e^{mx}$:
\begin{align*}
\left(e^{mx}\right)'' + e^{mx} &= 0 \\
m^2 e^{mx} + e^{mx} &= 0 \\
(m^2 + 1) e^{mx} &= 0
\end{align*}
For the LHS to be $0$ for all values of $x$, the following must be true:
$$
m^2 + 1 = 0
$$
Or:
$$
m = \pm i
$$
Therefore, the following are solutions to the equation:
\begin{align*}
y_1 &= e^{ix} \\
y_2 &= e^{-ix}
\end{align*}
And the general solution is:
$$
y = C_1 e^{ix} + C_2 e^{-ix}
$$
Where $C_1$ and $C_2$ are constants.
For real solutions, remember that:
\begin{align*}
e^{ix} &= \cos x + i \sin x \\
e^{-ix} &= \cos x - i \sin x
\end{align*}
Plug and simplify to get:
$$
y = (C_1 + C_2)\cos x + (C_1 i - C_2 i)\sin x
$$
And replace with real constants $B_1$ and $B_2$:
$$
y = B_1 \cos x + B_2 \sin x
$$
A: Guess a solution of $e^{rx}$. Plugging in, we get
$$e^{rx}(r^2+1)=0$$
$e^{rx}$ is never zero, meaning that the solutions are when $r^2+1=0$. Use that to find the valid values of r, and use the superposition principle (since it's a linear homogenous equation) to find the general solution.
EDIT:
$r^2+1=0$ gives $r=i,-i$. So by this and the superposition principle, your general solution is
$$Ae^{ix}+Be^{-ix}$$
But this is just one way of representing the general solution. All you need are two linearly independent solutions (which, for two solutions, just means one isn't a multiple of the other). If you know euler's formula, you can write
$$\cos(x)=\frac{e^{ix}+e^{-ix}} 2$$
$$\sin(x)=i\frac{e^{ix}-e^{-ix}} 2$$
So one "new" solution (cos) is found by taking $A=\frac 1 2, B=\frac 1 2$. Another (sin) is found by taking $A=\frac i 2, B=-\frac i 2$. Since these are linearly independent solutions, they can also be the "basis" of our solution set, and so we can see that another way of writing the above solution is 
$$A\cos(x)+B\sin(x)$$
A: An other standard method is to look for solutions of the form 
$$u(x)=\sum_{n=0}^\infty a_nx^n$$
Then a formal calculation (this step must be justified for solutions we might find) shows that $$u''(x)=\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n$$
which leads to 
$$u''(x)+u(x)= \sum_{n=0}^\infty \left((n+2)(n+1)a_{n+2}+a_n\right)x^n=0$$
This can only happen if $(n+2)(n+1)a_{n+2}+a_n =0$ for all $n$. 
I hope you see the picture and how to continue.
A: A less rigorous method (at least without an accompanying discussion that is more substantial than the problem itself) which does not involve guessing the solution involves "factorizing" the LHS:
$$\left(\frac{d^2}{dx^2}+1\right)y=0$$
$$\left(\frac{d}{dx}-i\right)\left(\frac{d}{dx}+i\right)y=0$$
$$\frac{dy}{dx}-iy=0 \qquad \frac{dy}{dx}+iy=0$$
$$y_1=C_1e^{ix} \qquad y_2=С_2e^{-ix}$$
Finally, the one that does not involve characteristic equation. Multiply by $y'$
$$y'y''+yy'=0$$
$$\frac{1}{2}\left(\frac{d{(y'^2)}}{dx}+\frac{d{(y^2)}}{dx}\right)=0$$
which is directly integrated once and leads to a second-order equation
