How do I solve $y''+4y=0$?

$y''+4y=0$ is a second order differential equation. First, change the equation to

$r^2+4 = 0$

This equation will will have complex conjugate roots, so the final answer would be in the form of $e^{\alpha x}(c_1\sin(\beta x) + c_2 \cos (\beta x))$ where $\alpha$ equals the real part of the complex roots and $\beta$ equals the imaginary part of (one of) the complex roots.

We need to use the quadratic formula \begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array}

In this equation $a =1$, $b=0$, and $c =4$

\begin{array}{*{20}c} {x = \frac{{ -0 \pm \sqrt {0^2 - 4(1)(4)} }}{{2}}} \\ \end{array} \begin{array}{*{20}c} {x = \frac{{ -0 \pm \sqrt {-16} }}{{2}}} \\ \end{array} \begin{array}{*{20}c} {x = \frac{{\pm \sqrt {-16} }}{{2}}} \\ \end{array} Since we can't have negative signs in the square root, we have an imaginary number $i$.

\begin{array}{*{20}c} {x = \frac{{\pm \sqrt {16}i }}{{2}}} \\ \end{array} Take the square root \begin{array}{*{20}c} {x = \frac{{\pm 4i }}{{2}}} \\ \end{array} \begin{array}{*{20}c} {x = \pm 2i} \\ \end{array} Now let's bring this equation back.

$y = e^{\alpha x}(c_1 \sin(\beta x) + c_2 \cos (\beta x))$

$\alpha = 0$ and $\beta = 2$

So your answer would be

$y = e^{0x}(c_1\sin(2x) + c_2\cos (2x))$

Since $e^{0x}$ is $1$ because $0$ multiplied by $x$ is just $0$ and $e^{0}$ is $1$.

The final answer is $y = (c_1\sin(2x) + c_2\cos (2x))$.

Use a "test solution": $y=e^{\lambda x}$. Then the characteristic equation is $\lambda^2+4=0$. Can you proceed from here?