Linear Second Order ODE/y'' + y = 0/y(0) = 2, y'(0) = 3

Theorem
The second order ODE:
 * $(1): \quad y'' + y = 0$

with initial conditions:
 * $y \left({0}\right) = 2$
 * $y' \left({0}\right) = 3$

has the solution:
 * $y = 3 \sin x + 2 \cos x$

Proof
From Second Order ODE: $y'' + y = 0$, the general solution of $(1)$ is:
 * $y = C_1 \sin x + C_2 \cos x$

Differentiating $x$:
 * $y' = C_1 \cos x - C_2 \sin x$

Thus for the initial conditions:

and:

Hence the result.