Linear Second Order ODE/y'' - y = 0/Proof 1

Proof
Note that:

and so by inspection:
 * $y_1 = e^x$

is a particular solution of $(1)$.

$(1)$ is in the form:
 * $y'' + P \left({x}\right) y' + Q \left({x}\right) y = 0$

where:
 * $P \left({x}\right) = 0$
 * $Q \left({x}\right) = -1$

From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another:
 * $y_2 \left({x}\right) = v \left({x}\right) y_1 \left({x}\right)$

where:
 * $\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \, \mathrm d x} \, \mathrm d x$

is also a particular solution of $(1)$.

We have that:

Hence:

and so:

From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:


 * $y = C_1 \sin x + k \left({-\dfrac C 2 e^{-x} }\right)$

where $k$ is arbitrary.

Setting $C_2 = - \dfrac {k C} 2$ yields the result:
 * $y = C_1 e^x + C_2 e^{-x}$