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How would i go about solving the differential equation> $$y''+y=0$$ (Identify the auxiliary equation.)

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up vote 3 down vote accepted

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:

$$ m^2 + 1 = 0 $$


$$ m = \pm i $$

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 $B_2$:

$$ y = B_1 \cos x + B_2 \sin x $$

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Guess a solution of $e^{rx}$. Plugging in, we get


$e^{rx}$ 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.


$r^2+1=0$ gives $r=i,-i$. So by this and the superposition principle, your general solution is


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 $$A\cos(x)+B\sin(x)$$

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How would i put it in terms of sin and cos? – Grigor May 28 '12 at 19:37
Use Euler's identity: $e^{ix}=\cos x+i\sin x$ – Dennis Gulko May 28 '12 at 19:42

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.

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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

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