avewright/proofwiki-math huggingface dataset:
Theorem: Linear Second Order ODE/x^2 y'' - 2 x y' + 2 y = 0

The second order ODE:
:$(1): \quad x^2 y' ' - 2 x y' + 2 y = 0$
has the general solution:
:$y = C_1 x + C_2 x^2$
on any closed real interval which does not contain $0$.

Proof:
Consider the functions:
:$\map {y_1} x = x$
:$\map {y_2} x = x^2$
We have that:
{{begin-eqn}}
{{eqn | l = \frac \d {\d x} \, x
      | r = 1
      | c = Power Rule for Derivatives
}}
{{eqn | l = \frac \d {\d x} \, x^2
      | r = 2 x
      | c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \frac {\d^2} {\d x^2} \, x
      | r = 0
      | c = 
}}
{{eqn | l = \frac {\d^2} {\d x^2} \, x^2
      | r = 2
      | c =  
}}
{{end-eqn}}
Putting $x$ and $x^2$ into $(1)$ in turn:
{{begin-eqn}}
{{eqn | l = x^2 \cdot 0 - 2 x \cdot 1 + 2 x
      | r = 2 x - 2 x
      | c = 
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = x^2 \cdot 2 - 2 x \cdot 2 x + 2 x^2
      | r = 2 x^2 - 4 x^2 + 2 x^2
      | c = 
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
Hence it can be seen that:
{{begin-eqn}}
{{eqn | l = \map {y_1} x
      | r = x
}}
{{eqn | l = \map {y_2} x
      | r = x^2
}}
{{end-eqn}}
are particular solutions to $(1)$.
Calculating the Wronskian of $y_1$ and $y_2$:
{{begin-eqn}}
{{eqn | l = \map W {y_1, y_2}
      | r = \begin{vmatrix} x & x^2 \\ 1 & 2 x \end{vmatrix}
      | c = 
}}
{{eqn | r = 2 x^2 - x^2
      | c = 
}}
{{eqn | r = x^2
      | c = 
}}
{{end-eqn}}
So the Wronskian of $y_1$ and $y_2$ is zero only when $x = 0$.
Let $\Bbb I = \closedint a b$ be a closed real interval such that $0 \notin \Bbb I$.
Thus from Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent:
:$y_1$ and $y_2$ are linearly independent on $\Bbb I$.
We can manipulate $(1)$ is a homogeneous linear second order ODE in the form:
:$y' ' + \map P x y' + \map Q x y = 0$
where $\map P x = \dfrac {-2 x} {x^2}$ and $\map Q x = \dfrac 2 {x^2}$.
So by Real Rational 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:
::$y = C_1 x + C_2 x^2$
:on any closed real interval $\Bbb I$ which does not include $0$.
{{qed}}
Theorem: Linear Second Order ODE/x^2 y'' - 2 x y' + 2 y = 0

The second order ODE:
:$(1): \quad x^2 y' ' - 2 x y' + 2 y = 0$
has the general solution:
:$y = C_1 x + C_2 x^2$
on any closed real interval which does not contain $0$.

Proof:
It can be seen that $(1)$ is an instance of the Cauchy-Euler Equation:
:$x^2 y' ' + p x y' + q y = 0$
where:
:$p = -2$
:$q = 2$
By Conversion of Cauchy-Euler Equation to Constant Coefficient Linear ODE, this can be expressed as:
:$\dfrac {\d^2 y} {\d t^2} + \paren {p - 1} \dfrac {\d y} {\d t} + q y = 0$
by making the substitution:
:$x = e^t$
Hence it can be expressed as:
:$(2): \quad \dfrac {\d^2 y} {\d t^2} - \dfrac {\d y} {\d t} + 2 y = 0$
From Linear Second Order ODE: $y' ' - 3 y' + 2 y = 0$, this has the general solution:
{{begin-eqn}}
{{eqn | l = y
      | r = C_1 e^t + C_2 e^{2 t}
      | c = 
}}
{{eqn | r = C_1 x + C_2 x^2
      | c = substituting $x = e^t$
}}
{{end-eqn}}
{{qed}}
Note that when $x = 0$, $e^t = x$ has no solution for $t$, and so is excluded from this solution.
Theorem: Linear Second Order ODE/x^2 y'' - 2 x y' + 2 y = 0/y(1) = 3, y'(1) = 5

The second order ODE:
:$(1): \quad x^2 y'' - 2 x y' + 2 y = 0$
with initial conditions:
{{begin-eqn}}
{{eqn | l = \map y 1
      | r = 3
}}
{{eqn | l = \map {y'} 1
      | r = 5
}}
{{end-eqn}}
has the particular solution:
:$y = x + 2 x^2$

Proof:
From Linear Second Order ODE: $x^2 y'' - 2 x y' + 2 y = 0$, the general solution of $(1)$ is:
:$y = C_1 x + C_2 x^2$
Differentiating {{WRT|Differentiation}} $x$:
:$y' = C_1 + 2 C_2 x$
Thus for the initial conditions:
{{begin-eqn}}
{{eqn | l = \map y 1
      | r = C_1 \times 1 + C_2 \times 1^2
      | c = 
}}
{{eqn | r = 3
      | c = 
}}
{{eqn | n = 2
      | ll= \leadsto
      | l = C_1 + C_2
      | r = 3
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \map {y'} 1
      | r = C_1 + 2 C_2 \times 1
      | c = 
}}
{{eqn | r = 5
      | c = 
}}
{{eqn | n = 3
      | ll= \leadsto
      | l = C_1 + 2 C_2
      | r = 5
      | c = 
}}
{{eqn | ll= \leadsto
      | l = C_2
      | r = 2
      | c = $(3) - (2)$
}}
{{eqn | ll= \leadsto
      | l = C_1 + 2
      | r = 3
      | c = substituting for $C_2$ in $(2)$
}}
{{eqn | ll= \leadsto
      | l = C_1
      | r = 1
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Linear Second Order ODE/y'' - 3 y' + 2 y = 0

The second order ODE:
:$(1): \quad y' ' - 3 y' + 2 y = 0$
has the general solution:
:$y = C_1 e^x + C_2 e^{2 x}$

Proof:
Consider the functions:
:$\map {y_1} x = e^x$
:$\map {y_2} x = e^{2 x}$
We have that:
{{begin-eqn}}
{{eqn | l = \frac {\d} {\d x} \, e^x
      | r = e^x
      | c = Power Rule for Derivatives
}}
{{eqn | l = \frac {\d} {\d x} \, e^{2 x}
      | r = 2 e^{2 x}
      | c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \frac {\d^2} {\d x^2} \, e^x
      | r = e^x
      | c = 
}}
{{eqn | l = \frac {\d^2} {\d x^2} \, e^{2 x}
      | r = 4 e^{2 x}
      | c =  
}}
{{end-eqn}}
Putting $e^x$ and $e^{2 x}$ into $(1)$ in turn:
{{begin-eqn}}
{{eqn | l = e^0 - 3 e^0 + 2 e^0
      | r = 1 - 3 + 2
      | c = 
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 4 e^{2 x} - 3 \times 2 e^{2 x} + 2 e^{2 x}
      | r = 4 - 6 + 2
      | c = 
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
Hence it can be seen that:
{{begin-eqn}}
{{eqn | l = \map {y_1} x
      | r = e^x
}}
{{eqn | l = \map {y_2} x
      | r = e^{2 x}
}}
{{end-eqn}}
are particular solutions to $(1)$.
Calculating the Wronskian of $y_1$ and $y_2$:
{{begin-eqn}}
{{eqn | l = \map W {y_1, y_2}
      | r = \begin {vmatrix} e^x & e^{2 x} \\ e^x & 2 e^{2 x} \end {vmatrix}
      | c = 
}}
{{eqn | r = e^x \times 2 e^{2 x} - e^{2 x} \times e^x
      | c = 
}}
{{eqn | r = 2 e^{3 x} - e^{3 x}
      | c = 
}}
{{eqn | r = e^{3 x}
      | c = 
}}
{{end-eqn}}
So the Wronskian of $y_1$ and $y_2$ is never zero.
Thus from Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent:
:$y_1$ and $y_2$ are linearly independent everywhere on $\R$.
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 $\map P x = -3$ and $\map Q x = 2$.
So by Constant Real Function is Continuous:
:$P$ and $Q$ are continuous on $\R$.
Thus from Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
:$(1)$ has the general solution:
::$y = C_1 e^x + C_2 e^{2 x}$
{{qed}}
Theorem: Linear Second Order ODE/y'' - 3 y' + 2 y = 0

The second order ODE:
:$(1): \quad y' ' - 3 y' + 2 y = 0$
has the general solution:
:$y = C_1 e^x + C_2 e^{2 x}$

Proof:
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 - 3 m + 2 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = 1$
:$m_2 = 2$
These are real and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = C_1 e^x + C_2 e^{2 x}$
{{qed}}
Theorem: Linear Second Order ODE/y'' - 3 y' + 2 y = 0/y(0) = -1, y'(0) = 1

The second order ODE:
:$(1): \quad y' ' - 3 y' + 2 y = 0$
with initial conditions:
:$\map y 0 = -1$
:$\map {y'} 0 = 1$
has the particular solution:
:$y = -3 e^x + 2 e^{2 x}$

Proof:
From Linear Second Order ODE: $y' ' - 3 y' + 2 y = 0$, the general solution of $(1)$ is:
:$y = C_1 e^x + C_2 e^{2 x}$
Differentiating {{WRT|Differentiation}} $x$:
:$y' = C_1 e^x + 2 C_2 e^{2 x}$
Thus for the initial conditions:
{{begin-eqn}}
{{eqn | l = \map y 0
      | r = C_1 e^0 + C_2 e^{2 \times 0}
      | c = 
}}
{{eqn | r = -1
      | c = 
}}
{{eqn | n = 2
      | ll= \leadsto
      | l = C_1 + C_2
      | r = -1
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \map {y'} 1
      | r = C_1 e^0 + 2 C_2 e^{2 \times 0}
      | c = 
}}
{{eqn | r = 1
      | c = 
}}
{{eqn | n = 3
      | ll= \leadsto
      | l = C_1 + 2 C_2
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = C_2
      | r = 2
      | c = $(3) - (2)$
}}
{{eqn | ll= \leadsto
      | l = C_1 + 2
      | r = -1
      | c = substituting for $C_2$ in $(2)$
}}
{{eqn | ll= \leadsto
      | l = C_1
      | r = -3
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Linear Second Order ODE/y'' - 4 y' + 4 y = 0

The second order ODE:
:$(1): \quad y' ' - 4 y' + 4 y = 0$
has the general solution:
:$y = C_1 e^{2 x} + C_2 x e^{2 x}$

Proof:
Consider the functions:
:$\map {y_1} x = e^{2 x}$
:$\map {y_2} x = x e^{2 x}$
We have that:
{{begin-eqn}}
{{eqn | l = \frac \d {\d x} \, e^{2 x}
      | r = 2 e^{2 x}
      | c = Power Rule for Derivatives
}}
{{eqn | l = \frac \d {\d x} \, x e^{2 x}
      | r = 2 x e^{2 x} + e^{2 x}
      | c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \frac {\d^2} {\d x^2} \, e^{2 x}
      | r = 4 e^{2 x}
      | c = 
}}
{{eqn | l = \frac {\d^2} {\d x^2} \, e^{2 x}
      | r = 4 x e^{2 x} + 2 e^{2 x} + 2 e^{2 x}
      | c =  
}}
{{eqn | r = 4 x e^{2 x} + 4 e^{2 x}
      | c =  
}}
{{end-eqn}}
Putting $e^{2 x}$ and $x e^{2 x}$ into $(1)$ in turn:
{{begin-eqn}}
{{eqn | l = 4 e^0 - 4 \times 2 e^0 + 4 e^0
      | r = 8 - 8 + 4
      | c = 
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 4 \times 0 e^0 + 4 e^0 - 4 \paren {2 \times 0 e^0 + e^0} + 4 \paren {0 e^0}
      | r = 0 + 4 - 4 + 0
      | c = 
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
Hence it can be seen that:
{{begin-eqn}}
{{eqn | l = \map {y_1} x
      | r = e^{2 x}
}}
{{eqn | l = \map {y_2} x
      | r = x e^{2 x}
}}
{{end-eqn}}
are particular solutions to $(1)$.
Calculating the Wronskian of $y_1$ and $y_2$:
{{begin-eqn}}
{{eqn | l =\map  W {y_1, y_2}
      | r = \begin{vmatrix} e^{2 x} & x e^{2 x} \\ 2 e^{2 x} & \paren {2 x + 1} e^{2 x} \end{vmatrix}
      | c = 
}}
{{eqn | r = e^{2 x} \times \paren {2 x + 1} e^{2 x} - x e^{2 x} \times 2 e^{2 x}
      | c = 
}}
{{eqn | r = \paren {2 x + 1} e^{4 x} - 2 x e^{4 x}
      | c = 
}}
{{eqn | r = e^{4 x}
      | c = 
}}
{{end-eqn}}
So the Wronskian of $y_1$ and $y_2$ is never zero.
Thus from Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent:
:$y_1$ and $y_2$ are linearly independent everywhere on $\R$.
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 $\map P x = -4$ and $\map Q x = 4$.
So by Constant Real Function is Continuous:
:$P$ and $Q$ are continuous on $\R$.
Thus from Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
:$(1)$ has the general solution:
::$y = C_1 e^{2 x} + C_2 x e^{2 x}$
{{qed}}
Theorem: Linear Second Order ODE/y'' - 4 y' + 4 y = 0

The second order ODE:
:$(1): \quad y' ' - 4 y' + 4 y = 0$
has the general solution:
:$y = C_1 e^{2 x} + C_2 x e^{2 x}$

Proof:
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 - 4 m + 4 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = m_2 = 2$
These are real and equal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = C_1 e^{2 x} + C_2 x e^{2 x}$
{{qed}}
Theorem: Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another

Let $\map {y_1} x$ be a particular solution to the homogeneous linear second order ODE:
:$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
such that $y_1$ is not the trivial solution.
Then there exists a standard procedure to determine another particular solution $\map {y_2} x$ of $(1)$ such that $y_1$ and $y_2$ are linearly independent.

Proof:
Let $\map {y_1} x$ be a non-trivial particular solution to $(1)$.
Thus, for all $C \in \R$, $C y_1$ is also a non-trivial particular solution to $(1)$.
Let $\map v x$ be a function of $x$ such that:
:$(2): \quad \map {y_2} x = \map v x \, \map {y_1} x$
is a particular solution to $(1)$ such that $y_1$ and $y_2$ are linearly independent.
Thus:
:$(3): \quad {y_2}' ' + \map P x {y_2}' + \map Q x y = 0$
Differentiating $(2)$ twice {{WRT|Differentiation}} $x$:
:${y_2}' = \map {v'} x \, \map {y_1} x + \map v x \, \map { {y_1}'} x$
and:
:${y_2}' ' = \map {v' '} x \, \map {y_1} x + 2 \, \map {v'} x \, \map { {y_1}'} x + \map v x \, \map { {y_1}' '} x$
Substituting these into $(3)$ and rearranging:
:$v \paren { {y_1}' ' + P {y_1}' + Q y_1} + v' ' y_1 + v' \paren {2 {y_1}' ' + P y_1} = 0$
Since $y_1$ is a particular solution of $(1)$, this reduces to:
:$v' ' y_1 + v' \paren {2 {y_1}' ' + P y_1} = 0$
which can be expressed as:
:$\dfrac {v' '} {v'} = -2 \dfrac { {y_1}'} {y_1} - P$
Integration gives:
:$\ds \ln v' = - 2 \ln y_1 - \int P \rd x$
This leads to:
:$v' = \dfrac 1 { {y_1}^2} e^{-\int P \rd x}$
and so:
:$\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
From Induced Solution to Homogeneous Linear Second Order ODE is Linearly Independent with Inducing Solution, $y_1$ and $v y_1$ are linearly independent.
{{qed}}
Theorem: Induced Solution to Homogeneous Linear Second Order ODE is Linearly Independent with Inducing Solution

Let $\map {y_1} x$ be a particular solution to the homogeneous linear second order ODE:
:$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
such that $y_1$ is not the trivial solution.
Let $\map {y_2} x$ be the real function defined as:
:$\map {y_2} x = \map v x \map {y_1} x$
where:
:$\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
Then $y_2$ and $y_1$ are linearly independent.

Proof:
This will be demonstrated by calculating the Wronskian of $y_1$ and $y_2$ and demonstrating that it is non-zero everywhere.
First we take the derivative of $v$ {{WRT|Differentiation}} $x$:
:$v' = \dfrac 1 { {y_1}^2} e^{- \int P \rd x}$
{{begin-eqn}}
{{eqn | l = \map W {y_1, y_2}
      | r = y_1 {y_2}' - y_2 {y_1}'
      | c = {{Defof|Wronskian}}
}}
{{eqn | r = y_1 \paren {v y_1}' - v y_1 {y_1}'
      | c = Definition of $y_2$
}}
{{eqn | r = y_1 \paren {v' y_1 + v {y_1}'} - v y_1 {y_1}'
      | c = Product Rule for Derivatives
}}
{{eqn | r = {y_1}^2 v' 
      | c = simplifying
}}
{{eqn | r = {y_1}^2 \dfrac 1 { {y_1}^2} e^{-\int P \rd x}
      | c = substituting for $v'$
}}
{{eqn | r = e^{-\int P \rd x}
      | c = simplifying
}}
{{end-eqn}}
As $\ds -\int P \rd x$ is a real function, $e^{-\int P \rd x}$ is non-zero wherever it is defined.
Hence from Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent, $y_1$ and $y_2$ are linearly independent.
{{qed}}
Theorem: Linear Second Order ODE/x^2 y'' + x y' - y = 0

The second order ODE:
:$(1): \quad x^2 y'' + x y' - y = 0$
has the general solution:
:$y = C_1 x + \dfrac {C_2} x$

Proof:
The particular solution:
:$y_1 = x$
can be found by inspection.
Let $(1)$ be written as:
:$(2): \quad y'' + \dfrac {y'} x - \dfrac y {x^2} = 0$
which is in the form:
:$y'' + \map P x y' + \map Q x y = 0$
where:
:$\map P x = \dfrac 1 x$
:$\map Q x = \dfrac 1 {x^2}$
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:
{{begin-eqn}}
{{eqn | l = \int P \rd x
      | r = \int \frac 1 x \rd x
      | c = 
}}
{{eqn | r = \ln x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = e^{-\int P \rd x}
      | r = e^{-\ln x}
      | c = 
}}
{{eqn | r = \frac 1 x
      | c =  
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = v
      | r = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x
      | c = Definition of $v$
}}
{{eqn | r = \int \dfrac 1 {x^2} \frac 1 x \rd x
      | c = 
}}
{{eqn | r = \int \dfrac {\d x} {x^3}
      | c = 
}}
{{eqn | r = -\dfrac 1 {2 x^2}
      | c = 
}}
{{end-eqn}}
and so:
{{begin-eqn}}
{{eqn | l = y_2
      | r = v y_1
      | c = Definition of $y_2$
}}
{{eqn | r = x \paren {-\dfrac 1 {2 x^2} }
      | c = 
}}
{{eqn | r = -\dfrac 1 {2 x}
      | c = 
}}
{{end-eqn}}
From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
:$y = C_1 x + k \paren {-\dfrac 1 {2 x} }$
where $k$ is arbitrary.
Setting $C_2 = -\dfrac k 2$ yields the result:
:$y = C_1 x + \dfrac {C_2} x$
{{qed}}
Theorem: Linear Second Order ODE/x^2 y'' + x y' - y = 0

The second order ODE:
:$(1): \quad x^2 y'' + x y' - y = 0$
has the general solution:
:$y = C_1 x + \dfrac {C_2} x$

Proof:
It can be seen that $(1)$ is an instance of the Cauchy-Euler Equation:
:$x^2 y'' + p x y' + q y = 0$
where:
:$p = 1$
:$q = -1$
By Conversion of Cauchy-Euler Equation to Constant Coefficient Linear ODE, this can be expressed as:
:$\dfrac {\d^2 y} {\d t^2} + \paren {p - 1} \dfrac {\d y} {\d t^2} + q y = 0$
by making the substitution:
:$x = e^t$
Hence it can be expressed as:
:$(2): \quad \dfrac {\d^2 y} {\d t^2} - y = 0$
From Linear Second Order ODE: $y'' - y = 0$, this has the general solution:
{{begin-eqn}}
{{eqn | l = y
      | r = C_1 e^t + C_2 e^{-t}
      | c = 
}}
{{eqn | r = C_1 x + C_2 x^{-1}
      | c = substituting $x = e^t$
}}
{{eqn | r = C_1 x + \frac {C_2} x
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Linear Second Order ODE/y'' - y = 0

The second order ODE:
:$(1): \quad y' ' - y = 0$
has the general solution:
:$y = C_1 e^x + C_2 e^{-x}$

Proof:
Note that:
{{begin-eqn}}
{{eqn | l = y_1
      | r = e^x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y'
      | r = e^x
      | c = Derivative of Exponential Function
}}
{{eqn | ll= \leadsto
      | l = y' '
      | r = e^x
      | c = Derivative of Exponential Function
}}
{{end-eqn}}
and so by inspection:
:$y_1 = e^x$
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:
{{begin-eqn}}
{{eqn | l = \int P \rd x
      | r = \int 0 \rd x
      | c = 
}}
{{eqn | r = k
      | c = where $k$ is arbitrary
}}
{{eqn | ll= \leadsto
      | l = e^{-\int P \rd x}
      | r = e^{-k}
      | c = 
}}
{{eqn | r = C
      | c =  where $C$ is also arbitrary
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = v
      | r = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x
      | c = Definition of $v$
}}
{{eqn | r = \int \dfrac 1 {e^{2 x} } C \rd x
      | c = 
}}
{{eqn | r = \int C e^{-2 x} \rd x
      | c = 
}}
{{eqn | r = -\frac {C e^{-2 x} } 2
      | c = Primitive of $e^{a x}$
}}
{{end-eqn}}
and so:
{{begin-eqn}}
{{eqn | l = y_2
      | r = v y_1
      | c = Definition of $y_2$
}}
{{eqn | r = e^x \paren {-\frac C 2 e^{-2 x} }
      | c = 
}}
{{eqn | r = -\frac C 2 e^{-x}
      | c = 
}}
{{end-eqn}}
From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
:$y = C_1 \sin x + k \paren {-\dfrac C 2 e^{-x} }$
where $k$ is arbitrary.
Setting $C_2 = - \dfrac {k C} 2$ yields the result:
:$y = C_1 e^x + C_2 e^{-x}$
{{qed}}
Theorem: Linear Second Order ODE/y'' - y = 0

The second order ODE:
:$(1): \quad y' ' - y = 0$
has the general solution:
:$y = C_1 e^x + C_2 e^{-x}$

Proof:
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 - 1 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = 1$
:$m_2 = -1$
These are real and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = C_1 e^x + C_2 e^{- x}$
{{qed}}
Theorem: Linear Second Order ODE/y'' - y = 0

The second order ODE:
:$(1): \quad y' ' - y = 0$
has the general solution:
:$y = C_1 e^x + C_2 e^{-x}$

Proof:
This is an instance of:
:Linear Second Order ODE: $y' ' - k^2 y = 0$
which yields:
:$y = C_1 e^{k x} + C_2 e^{-k x}$
where $k = 1$.
Hence the result.
{{qed}}
Theorem: Second Order ODE/x y'' + 3 y' = 0

The second order ODE:
:$(1): \quad x y' ' + 3 y' = 0$
has the general solution:
:$y = C_1 + \dfrac {C_2} {x^2}$

Proof:
Note that:
{{begin-eqn}}
{{eqn | l = y_1
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y'
      | r = 0
      | c = Derivative of Constant
}}
{{eqn | ll= \leadsto
      | l = y' '
      | r = 0
      | c = Derivative of Constant
}}
{{end-eqn}}
and so by inspection:
:$y_1 = 1$
is a particular solution of $(1)$.
$(1)$ can be expressed as:
:$(2): \quad y' ' + \dfrac 3 x y' = 0$
which is in the form:
:$y' ' + \map P x y' + \map Q x y = 0$
where:
:$\map P x = \dfrac 3 x$
:$\map Q x = 0$
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:
{{begin-eqn}}
{{eqn | l = \int P \rd x
      | r = \int \dfrac 3 x \rd x
      | c = 
}}
{{eqn | r = 3 \ln x
      | c = Primitive of Reciprocal
}}
{{eqn | r = \ln x^3
      | c = 
}}
{{eqn | ll= \leadsto
      | l = e^{-\int P \rd x}
      | r = e^{-\ln x^3}
      | c = 
}}
{{eqn | r = \frac 1 {x^3}
      | c =  
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = v
      | r = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x
      | c = Definition of $v$
}}
{{eqn | r = \int \frac 1 {x^3} \rd x
      | c = 
}}
{{eqn | r = -\frac 1 {2 x^2}
      | c = 
}}
{{end-eqn}}
and so:
{{begin-eqn}}
{{eqn | l = y_2
      | r = v y_1
      | c = Definition of $y_2$
}}
{{eqn | r = -\frac 1 {2 x^2}
      | c = 
}}
{{end-eqn}}
From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
:$y = C_1 + k \paren {-\frac 1 {2 x^2} }$
where $k$ is arbitrary.
Setting $C_2 = -\dfrac k 2$ yields the result:
:$y = C_1 + \dfrac {C_2} {x^2}$
{{qed}}
</onlyinclude>
Theorem: Linear Second Order ODE/x^2 y'' + x y' - 4 y = 0

The second order ODE:
:$(1): \quad x^2 y'' + x y' - 4 y = 0$
has the general solution:
:$y = C_1 x^2 + \dfrac {C_2} {x^2}$

Proof:
Note that:
{{begin-eqn}}
{{eqn | l = y_1
      | r = x^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y'
      | r = 2 x
      | c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = y''
      | r = 2
      | c = Power Rule for Derivatives
}}
{{end-eqn}}
and so by inspection:
:$y_1 = x^2$
is a particular solution of $(1)$.
$(1)$ can be expressed as:
:$(2): \quad y'' + \dfrac 1 x y' - \dfrac 4 {x^2} y = 0$
which is in the form:
:$y'' + \map P x y' + \map Q x y = 0$
where:
:$\map P x = \dfrac 1 x$
:$\map Q x = -\dfrac 4 {x^2}$
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:
{{begin-eqn}}
{{eqn | l = \int P \rd x
      | r = \int \dfrac 1 x \rd x
      | c = 
}}
{{eqn | r = \ln x
      | c = Primitive of Reciprocal
}}
{{eqn | ll= \leadsto
      | l = e^{-\int P \rd x}
      | r = e^{-\ln x}
      | c = 
}}
{{eqn | r = \frac 1 x
      | c =  
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = v
      | r = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x
      | c = Definition of $v$
}}
{{eqn | r = \int \dfrac 1 {x^4} \frac 1 x \rd x
      | c = 
}}
{{eqn | r = \int \dfrac 1 {x^5} \rd x
      | c = 
}}
{{eqn | r = - \frac 1 {4 x^4}
      | c = 
}}
{{end-eqn}}
and so:
{{begin-eqn}}
{{eqn | l = y_2
      | r = v y_1
      | c = Definition of $y_2$
}}
{{eqn | r = -\frac 1 {4 x^4} x^2
      | c = 
}}
{{end-eqn}}
From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
:$y = C_1 x^2 + k \paren {-\dfrac 1 {4 x^2} }$
where $k$ is arbitrary.
Setting $C_2 = - \dfrac k 4$ yields the result:
:$y = C_1 x^2 + \dfrac {C_2} {x^2}$
{{qed}}