Linear Second Order ODE/y'' = y'

From ProofWiki
Jump to navigation Jump to search

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
\(\ds \) \(=\) \(\ds -e^{-x} \paren {x^2 + 2 x + 2} + C\)
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds A_1 e^x + A_2\)

$\blacksquare$


Sources