Linear Second Order ODE/y'' + y = 0

From ProofWiki
Jump to navigation Jump to search

Theorem

The second order ODE:

$(1): \quad y'' + y = 0$

has the general solution:

$y = C_1 \sin x + C_2 \cos x$


Proof 1

Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as:

$p \dfrac {\d p} {\d y} = -y$

where $p = \dfrac {\d y} {\d x}$.

From:

First Order ODE: $y \rd y = k x \rd x$

with $k = 1$, this has the solution:

$p^2 = -y^2 + C$

or:

$p^2 + y^2 = C$

As the left hand side is the sum of squares, $C$ has to be positive for this to have any solutions.


Thus, let $C = \alpha^2$.

Then:

\(\ds p^2 + y^2\) \(=\) \(\ds \alpha^2\)
\(\ds \leadsto \ \ \) \(\ds p = \dfrac {\d y} {\d x}\) \(=\) \(\ds \pm \sqrt {\alpha^2 - y^2}\)
\(\ds \leadsto \ \ \) \(\ds \int \dfrac {\d y} {\sqrt {\alpha^2 - y^2} }\) \(=\) \(\ds \int \pm 1 \rd x\) Solution to Separable Differential Equation
\(\ds \leadsto \ \ \) \(\ds \arcsin \dfrac y \alpha\) \(=\) \(\ds \pm x + \beta\) Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds \alpha \map \sin {\pm x + \beta}\)
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds A \map \sin {x + B}\)

From Multiple of Sine plus Multiple of Cosine: Sine Form, this can be expressed as:

$y = C_1 \sin x + C_2 \cos x$

$\blacksquare$


Proof 2

$(1)$ can be seen to be a special case of:

$(2): \quad$ Linear Second Order ODE: $y'' + k^2 y = 0$

with $k = 1$.

$(2)$ has the solution:

$y = C_1 \sin k x + C_2 \cos k x$

Hence setting $k = 1$:

$y = C_1 \sin x + C_2 \cos x$

$\blacksquare$


Proof 3

We have that:

\(\ds \frac \d {\d x} \, \sin x\) \(=\) \(\ds \cos x\) Derivative of Sine Function
\(\ds \frac \d {\d x} \, \cos x\) \(=\) \(\ds -\sin x\) Derivative of Cosine Function
\(\ds \leadsto \ \ \) \(\ds \frac {\d^2} {\d x^2} \, \cos x\) \(=\) \(\ds -\cos x\)
\(\ds \frac {\d^2} {\d x^2} \, \sin x\) \(=\) \(\ds -\sin x\)


Hence it can be seen by inspection that:

\(\ds \map {y_1} x\) \(=\) \(\ds \sin x\)
\(\ds \map {y_2} x\) \(=\) \(\ds \cos x\)

are particular solutions to $(1)$.

It is noted that:

$\dfrac {y_1} {y_2} = \tan x$

which is not a constant function on any closed interval $\closedint a b$.


Calculating the Wronskian of $y_1$ and $y_2$:

\(\ds \map W {y_1, y_2}\) \(=\) \(\ds \begin{vmatrix} \sin x & \cos x \\ \cos x & -\sin x \end{vmatrix}\)
\(\ds \) \(=\) \(\ds -\sin^2 x - \cos^2 x\)
\(\ds \) \(=\) \(\ds -1\) Sum of Squares of Sine and Cosine

So the Wronskian of $y_1$ and $y_2$ is never zero.


We have that $(1)$ is a homogeneous linear second order ODE in the form:

$y'' + \map P x y' + \map Q x y = 0$

where $\map P x = 0$ and $\map Q x$ are constant functions.

So by Constant Real Function is Continuous: $P$ and $Q$ are continuous on $\closedint a b$

Thus from Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:

$(1)$ has the general solution:
$y = C_1 \sin x + C_2 \cos x$
on $\closedint a b$.

Since $\closedint a b$ can be extended indefinitely without introducing points where $P$ or $Q$ are discontinuous, the general solution is valid for all $x$.

$\blacksquare$


Proof 4

Note that:

\(\ds y_1\) \(=\) \(\ds \sin x\)
\(\ds \leadsto \ \ \) \(\ds y'\) \(=\) \(\ds \cos x\) Derivative of Sine Function
\(\ds \leadsto \ \ \) \(\ds y''\) \(=\) \(\ds -\sin x\) Derivative of Cosine Function

and so:

$y_1 = x$

is a particular solution of $(1)$.


$(1)$ is in the form:

$y'' + \map P x y' + \map Q x y = 0$

where:

$\map P x = 0$
$\map Q x = 1$


From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another:

$\map {y_2} x = \map v x \, \map {y_1} x$

where:

$\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$

is also a particular solution of $(1)$.


We have that:

\(\ds \int P \rd x\) \(=\) \(\ds \int 0 \rd x\)
\(\ds \) \(=\) \(\ds k\) where $k$ is arbitrary
\(\ds \leadsto \ \ \) \(\ds e^{-\int P \rd x}\) \(=\) \(\ds e^{-k}\)
\(\ds \) \(=\) \(\ds C\) where $C$ is also arbitrary


Hence:

\(\ds v\) \(=\) \(\ds \int \dfrac 1 { {y_1}^2} e^{- \int P \rd x} \rd x\) Definition of $v$
\(\ds \) \(=\) \(\ds \int \dfrac 1 {\sin^2 x} C \rd x\)
\(\ds \) \(=\) \(\ds \int C \csc^2 x \rd x\) Definition of Cosecant
\(\ds \) \(=\) \(\ds -C \cot x\) Primitive of $\csc^2 x$


and so:

\(\ds y_2\) \(=\) \(\ds v y_1\) Definition of $y_2$
\(\ds \) \(=\) \(\ds \sin x \paren {-C \cot x}\)
\(\ds \) \(=\) \(\ds -C \cos x\) Definition of Cotangent, and simplifying


From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:

$y = C_1 \sin x + k \paren {-C \cos x}$

where $k$ is arbitrary.


Setting $C_2 = - k C$ yields the result:

$y = C_1 \sin x + C_2 \cos x$

$\blacksquare$


Proof 5

Taking Laplace transforms, we have:

$\laptrans {y'' + y} = \laptrans 0$

From Laplace Transform of Constant Mapping, we have:

$\laptrans 0 = 0$

We also have:

\(\ds \laptrans {y'' + y}\) \(=\) \(\ds \laptrans {y''} + \laptrans y\) Linear Combination of Laplace Transforms
\(\ds \) \(=\) \(\ds s^2 \laptrans y - s \map y 0 - \map {y'} 0 + \laptrans y\) Laplace Transform of Second Derivative

So:

$\paren {s^2 + 1} \laptrans y = s \map y 0 + \map {y'} 0$

Giving:

$\laptrans y = \map y 0 \dfrac s {s^2 + 1} + \map {y'} 0 \dfrac 1 {s^2 + 1}$

So:

\(\ds y\) \(=\) \(\ds \invlaptrans {\map y 0 \frac s {s^2 + 1} + \map {y'} 0 \frac 1 {s^2 + 1} }\) Definition of Inverse Laplace Transform
\(\ds \) \(=\) \(\ds \map y 0 \invlaptrans {\frac s {s^2 + 1} } + \map {y'} 0 \invlaptrans {\frac 1 {s^2 + 1} }\) Linear Combination of Laplace Transforms
\(\ds \) \(=\) \(\ds \map y 0 \invlaptrans {\laptrans {\cos x} } + \map {y'} 0 \invlaptrans {\laptrans {\sin x} }\) Laplace Transform of Cosine, Laplace Transform of Sine
\(\ds \) \(=\) \(\ds \map y 0 \cos x + \map {y'} 0 \sin x\) Definition of Inverse Laplace Transform

Setting $C_1 = \map {y'} 0$ and $C_2 = \map y 0$ gives the result.

$\blacksquare$