# Linear Second Order ODE/y'' = y'

## Theorem

The second order ODE:

$(1): \quad y'' = y'$

has the general solution:

$y = A_1 e^x + A_2$

## Proof 1

The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable.

Substitute $p$ for $y'$ in $(1)$:

 $\ds \dfrac {\d p} {\d x}$ $=$ $\ds p$ where $p = \dfrac {\d y} {\d x}$ $\ds \leadsto \ \$ $\ds \int \rd x$ $=$ $\ds \int \frac {\d p} p$ Separation of Variables $\ds \leadsto \ \$ $\ds x$ $=$ $\ds \ln p + C$ Primitive of $\dfrac 1 x$ $\ds \leadsto \ \$ $\ds p = \frac {\d y} {\d x}$ $=$ $\ds A_1 e^x$ where $A_1 = e^C$ $\ds \leadsto \ \$ $\ds \int \rd y$ $=$ $\ds \int A_1 e^x \rd x$ Separation of Variables $\ds \leadsto \ \$ $\ds y$ $=$ $\ds A_1 e^x + A_2$ Primitive of Exponential Function

$\blacksquare$

## Proof 2

Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as:

 $\ds p \frac {\d p} {\d y}$ $=$ $\ds p$ where $p = \dfrac {\d y} {\d x}$ $\ds \leadsto \ \$ $\ds \int \rd y$ $=$ $\ds \int \frac {p \rd p} p$ Separation of Variables $\ds \leadsto \ \$ $\ds \int \rd y$ $=$ $\ds \int \rd p$ $\ds \leadsto \ \$ $\ds y$ $=$ $\ds p + A_2$ Primitive of Constant $\ds \leadsto \ \$ $\ds p = \frac {\d y} {\d x}$ $=$ $\ds y - A_2$ $\text {(1)}: \quad$ $\ds \leadsto \ \$ $\ds \frac {\d y} {\d x} - y$ $=$ $\ds -A_2$

$(1)$ is a linear first order ODE in the form:

$\dfrac {\d y} {\d x} + \map P x y = \map Q x$

where:

$\map P x = -1$
$\map Q x = -A_2$

Thus:

 $\ds \int \map P x \rd x$ $=$ $\ds -\int \rd x$ $\ds$ $=$ $\ds -x$ $\ds \leadsto \ \$ $\ds e^{\int P \rd x}$ $=$ $\ds e^{-x}$

Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:

 $\ds \map {\dfrac {\d} {\d x} } {y e^{-x} }$ $=$ $\ds -A_2 e^{-x}$ $\ds \leadsto \ \$ $\ds y e^{-x}$ $=$ $\ds -A_2 \int e^ \rd x$ $\ds$ $=$ $\ds A_2 e^{-x} + A_1$ Primitive of Exponential Function $\ds$ $=$ $\ds -e^{-x} \paren {x^2 + 2 x + 2} + C$ $\ds \leadsto \ \$ $\ds y$ $=$ $\ds A_1 e^x + A_2$

$\blacksquare$