Solve the differential equation y'' + y = 0.

Step-by-step solution

1. 1

   Substitute the exponential trial solution $y=e^{rx}$. Then $y''=r^2e^{rx}$, and the equation becomes

   $r^2e^{rx}+e^{rx}=e^{rx}\left(r^2+1\right)=0$

2. 2

   Read off the characteristic equation. Since $e^{rx}\neq 0$ for all real $x$:

   $r^2+1=0\quad\Longrightarrow\quad r^2=-1\quad\Longrightarrow\quad r=\pm i$

3. 3

   Write the real general solution. With $\alpha=0$ and $\beta=1$ in the template $e^{\alpha x}\left(C_1\cos\beta x+C_2\sin\beta x\right)$:

   $y(x)=C_1\cos x+C_2\sin x$

   Two arbitrary constants is exactly right for a second-order equation — they are fixed by two initial conditions such as $y(0)$ and $y'(0)$.

4. 4

   Verify both basis solutions. For $y=\cos x$: $y''=-\cos x$, so $y''+y=0$ $\checkmark$. For $y=\sin x$: $y''=-\sin x$, again $0$ $\checkmark$.

5. 5

   Compare with $y''+4y=0$. That equation gives $r=\pm 2i$ and solutions in $\cos 2x,\sin 2x$: the coefficient of $y$ is the square of the angular frequency, so $y''+\omega^2y=0$ oscillates at $\omega$. Here $\omega=1$ and the period is $2\pi$.