Linear Second Order ODE/y'' + y = 0/Proof 3

Proof
We have that:

Hence it can be seen by inspection that:

are particular solutions to $(1)$.

It is noted that:
 * $\dfrac {y_1} {y_2} = \tan x$

which is not a constant function on any closed interval $\left[{a \,.\,.\, b}\right]$.

Calculating the Wronskian of $y_1$ and $y_2$:

So the Wronskian of $y_1$ and $y_2$ is never zero.

We have that $(1)$ is a homogeneous linear second order ODE in the form:
 * $y'' + P \left({x}\right) y' + Q \left({x}\right) y = 0$

where $P \left({x}\right) = 0$ and $Q \left({x}\right)$ are constant functions.

So by Constant Real Function is Continuous: $P$ and $Q$ are continuous on $\left[{a \,.\,.\, b}\right]$

Thus from Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
 * $(1)$ has the general solution:
 * $y = C_1 \sin x + C_2 \cos x$
 * on $\left[{a \,.\,.\, b}\right]$.

Since $\left[{a \,.\,.\, b}\right]$ can be extended indefinitely without introducing points where $P$ or $Q$ are discontinuous, the general solution is valid for all $x$.