Linear Second Order ODE/y'' + y = 0
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:
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.
$\blacksquare$