Linear Second Order ODE/y'' = y'
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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 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 $p = \dfrac {\d y} {\d x}$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \rd x\) | \(=\) | \(\ds \int \frac {\d p} p\) | 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 |
$\blacksquare$
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 $p = \dfrac {\d y} {\d x}$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \rd y\) | \(=\) | \(\ds \int \frac {p \rd p} p\) | Solution to Separable Differential Equation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \rd y\) | \(=\) | \(\ds \int \rd p\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds p + A_2\) | 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
- 1963: Morris Tenenbaum and Harry Pollard: Ordinary Differential Equations ... (previous) ... (next): Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation: $(3.14)$