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

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Theorem

The second order ODE:

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

has the general solution:

$y = A_1 e^x + A_2$


Proof

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

$\blacksquare$