Solve the differential equation y'' + 4y = 0.

Step-by-step solution

1. 1

   Try an exponential trial solution. For a linear ODE with constant coefficients, $y=e^{rx}$ always turns the equation into an algebraic one:

   $y=e^{rx}\Rightarrow y''=r^2e^{rx}$

2. 2

   Form the characteristic equation. Substituting and factoring out $e^{rx}$, which is never zero:

   $r^2e^{rx}+4e^{rx}=e^{rx}\left(r^2+4\right)=0\quad\Longrightarrow\quad r^2+4=0$

3. 3

   Solve for $r$.

   $r^2=-4\quad\Longrightarrow\quad r=\pm 2i$

   Purely imaginary roots signal pure oscillation with no growth or decay.

4. 4

   Convert the complex exponentials to real form. For roots $r=\alpha\pm\beta i$ the real general solution is $e^{\alpha x}\left(C_1\cos\beta x+C_2\sin\beta x\right)$. Here $\alpha=0$ and $\beta=2$, so the exponential envelope is $e^{0}=1$:

   $y(x)=C_1\cos 2x+C_2\sin 2x$

5. 5

   Verify by direct substitution. With $y=\cos 2x$: $y''=-4\cos 2x$, so $y''+4y=0$ $\checkmark$. With $y=\sin 2x$: $y''=-4\sin 2x$, again $0$ $\checkmark$. Because the equation is linear, any combination of the two also works. Physically this is simple harmonic motion of angular frequency $\omega=2$ and period $\pi$.