# What is the general solution of the differential equation  y'' +4y =0?

The differential equation $y ' ' + 4 y = 0$ is what we call second order differential equation.

So we need to find its characteristic equation which is

${r}^{2} + 4 = 0$

This equation will will have complex conjugate roots, so the final answer would be in the form of y=e^(αx)*(c_1*sin⁡(βx)+c_2*cos⁡(βx)) where α equals the real part of the complex roots and β equals the imaginary part of (one of) the complex roots.

We need to use the quadratic formula
r=[−b±sqrt(b^2−4ac)]/[2*a]

when $a \cdot {r}^{2} + b \cdot r + c = 0$

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

Hence the roots are ${r}_{1} = 2 i$ and ${r}_{2} = - 2 i$

Now the form of y=e^(αx)*(c_1*sin⁡(βx)+c_2*cos⁡(βx))

where $a = 0$ and β=2

becomes

$y = {e}^{0 \cdot x} \cdot \left({c}_{1} \cdot \sin \left(2 \cdot x\right) + {c}_{2} \cdot \cos \left(2 \cdot x\right)\right)$

Finally

$y = \left({c}_{1} \cdot \sin \left(2 \cdot x\right) + {c}_{2} \cdot \cos \left(2 \cdot x\right)\right)$

The coefficients ${c}_{1} , {c}_{2}$ can be determined if we have initial conditions for the differential equation.

Jun 15, 2017

$y = A \cos 2 x + B \sin 2 x$

#### Explanation:

We have:

$y ' ' + 4 y = 0$

This is a second order linear Homogeneous Differentiation Equation. The standard approach is to look at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, i.e.

${m}^{2} + 4 = 0$

This has two distinct complex solutions:

$m = \pm 2 i$, or $m = 0 \pm 2 i$

And so the solution to the DE is;

$\setminus \setminus \setminus \setminus \setminus y = {e}^{0} \left(A \cos 2 x + B \sin 2 x\right)$ Where $A , B$ are arbitrary constants
$\therefore y = A \cos 2 x + B \sin 2 x$