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

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 $p = \dfrac {\d y} {\d x}$

\(\ds \leadsto \ \ \)

\(\ds \int \rd x\)

\(=\)

\(\ds \int \frac {\d p} p\)

Solution to Separable Differential Equation

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds \ln p + C\)

Primitive of $\dfrac 1 x$

\(\ds \leadsto \ \ \)

\(\ds p = \frac {\d y} {\d x}\)

\(=\)

\(\ds A_1 e^x\)

where $A_1 = e^C$

\(\ds \leadsto \ \ \)

\(\ds \int \rd y\)

\(=\)

\(\ds \int A_1 e^x \rd x\)

Solution to Separable Differential Equation

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds A_1 e^x + A_2\)

Primitive of Exponential Function

$\blacksquare$