Linear Second Order ODE/y'' - k^2 y = 0/Proof 1
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Theorem
The second order ODE:
- $(1): \quad y - k^2 y = 0$
has the general solution:
- $y = C_1 e^{k x} + C_2 e^{-k x}$
Proof
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as:
\(\ds p \frac {\d p} {\d y}\) | \(=\) | \(\ds k^2 y\) | where $p = \dfrac {\d y} {\d x}$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds p^2\) | \(=\) | \(\ds k^2 y^2 + k^2 \alpha\) | First Order ODE: $y \rd y = k x \rd x$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds p = \dfrac {\d y} {\d x}\) | \(=\) | \(\ds \pm k \sqrt {y^2 + k^2 \alpha}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \dfrac {\d y} {\sqrt {y^2 + \alpha} }\) | \(=\) | \(\ds \int \pm k \rd x\) | Solution to Separable Differential Equation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \ln {y + \sqrt{y^2 + \alpha} }\) | \(=\) | \(\ds \pm k x + \beta\) | Primitive of $\dfrac 1 {\sqrt {x^2 + k} }$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds y + \sqrt {y^2 + \alpha}\) | \(=\) | \(\ds e^{\pm k x + \beta}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds C e^{\pm k x}\) | where $C = e^\beta$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds y^2 + \alpha\) | \(=\) | \(\ds \paren {C e^{\pm k x} - y}^2\) | |||||||||||
\(\ds \) | \(=\) | \(\ds C^2 e^{\pm 2 k x} - 2 C e^{\pm k x} + y^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \frac {C^2 e^{\pm 2 k x} - \alpha} {2 C e^{\pm k x} }\) | Quadratic Formula | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac {C e^{\pm k x} - \frac \alpha C e^{\mp k x} } 2\) | Quadratic Formula |
Setting $C_1 = \dfrac C 2$ and $C_2 = - \dfrac \alpha {2 C}$:
- $y = C_1 e^{\pm k x} + C_2 e^{\mp k x}$
which is the same thing as:
- $y = C_1 e^{k x} + C_2 e^{-k x}$
by allowing for the constants to be interchanged.
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.11$: Problem $1 \ \text{(c)}$