Second Order ODE/y'' = f(y)
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Theorem
The second order ODE:
- $\dfrac {\d^2 y} {\d x^2} = \map f y$
has the general solution:
- $\dfrac {\sqrt 2} 2 \ds \int \dfrac {\d y} {\sqrt {\ds \int \map f y \rd y + C_1} } = x + C_2$
where $C_1$ and $C_2$ are arbitrary constants.
Proof
\(\ds \dfrac {\d^2 y} {\d x^2}\) | \(=\) | \(\ds \map f y\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2 \dfrac {\d y} {\d x} \dfrac {\d^2 y} {\d x^2}\) | \(=\) | \(\ds 2 \map f y \dfrac {\d y} {\d x}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int 2 \dfrac {\d y} {\d x} \dfrac {\d^2 y} {\d x^2} \rd x\) | \(=\) | \(\ds \int 2 \map f y \dfrac {\d y} {\d x} \rd x\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {\dfrac {\d y} {\d x} }^2\) | \(=\) | \(\ds 2 \int \map f y \rd y + C\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {\d y} {\d x}\) | \(=\) | \(\ds \sqrt {2 \int \map f y \rd y + C_1}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \dfrac {\d y} {\sqrt {2 \ds \int \map f y \rd y + C_1} }\) | \(=\) | \(\ds \int \rd x\) | Solution to Separable Differential Equation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {\sqrt 2} 2 \int \dfrac {\d y} {\sqrt {\ds \int \map f y \rd y + C_1} }\) | \(=\) | \(\ds x + C_2\) |
$\blacksquare$
Sources
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): differential equation: differential equations of the second order: $(2)$ Equations of the form $\dfrac {\d^2 y} {\d x^2} = \map f y$
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): differential equation: differential equations of the second order: $(2)$ Equations of the form $\dfrac {\d^2 y} {\d x^2} = \map f y$