Linear Second Order ODE/y'' - k^2 y = 0

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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 1

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.

$\blacksquare$


Proof 2

It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.

Its auxiliary equation is:

$(2): \quad m^2 - k^2 = 0$

From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:

$m_1 = k$
$m_2 = -k$


These are real and unequal.

So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:

$y = C_1 e^{k x} + C_2 e^{-k x}$

$\blacksquare$


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