nyuuzyou/wikis huggingface dataset:
Linear Programming/Examples/Arbitrary Example 1

== Examples of Linear Programming ==
Let it be required to find the minimum of the function:
:$\map U {x, y} = 4 x + 3 y$
subject to the conditions:
The maximum feasible value of $U$ is given by:
:$U = 65$
== Proof ==
Let us present the above information in a graphical form:
:
The lines represent the boundary lines of the conditions:
:the line $AB$ represents the condition $x + y = 20$
:the line $CD$ represents the condition $3 x + y = 30$
while the axes represent the conditions $x = 0$ and $y = 0$.
Hence the region in which feasible solutions exist is the colored region $OAED$.
The dashed parallel lines represent the equation $U = 4 x + 3 y$ for various values of $U$.
As $U$ increases, the dashed line moves to the right.
Hence the optimum value of $U$ occurs when $U$ is chosen so that the line passes through the point $E$.
This is where the lines $x + y = 20$ and $3 x + y = 30$ intersect.
Solving these simultaneous equations gives us:
Hence the maximum feasible value of $U$ is given by:
which is the optimum solution.
== Sources ==
* 
* 
Category:Examples of Linear Programming
Linear Ratio Scale/Examples/Grams and Ounces

== Examples of Linear Ratio Scales ==
The scales for the measurement of weight in grams and ounces are linear ratio scales.
Let $w_1$ and $w_2$ be two weights measured as $10 \, \mathrm g$ and $20 \, \mathrm g$ on the gram scale.
It is noted that $\dfrac {w_2} {w_1} = 2$.
On the ounce scale, we have that $w_1$ and $w_2$ measure as $0 \cdotp 35 \, \mathrm {oz}$ and $0 \cdotp 70 \, \mathrm {oz}$ respectively.
As can be seen, once again $\dfrac {w_2} {w_1} = 2$.
== Sources ==
* 
Category:Examples of Linear Ratio Scales
Linear Second Order ODE/(1 + x^2) y'' + x y' = 0

== Theorem ==
The second order ODE:
:$\paren {1 + x^2} y + x y' = 0$
has the general solution:
:$y = C_1 \map \ln {x + \sqrt {x^2 + 1} } + C_2$
== Proof ==
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable.
Substitute $p$ for $y'$:
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $23$
Category:Examples of Homogeneous LSOODEs
Linear Second Order ODE/(x^2 + x) y'' + (2 - x^2) y' - (2 + x) y = 0

== Theorem ==
The second order ODE:
:$(1): \quad \paren {x^2 + x} y + \paren {2 - x^2} y' - \paren {2 + x} y = 0$
has the general solution:
:$y = C_1 e^x + \dfrac {C_2} x$
== Proof 1 ==
== Proof 2 ==
Category:Examples of Homogeneous LSOODEs
Category:Linear Second Order ODE/(x^2 + x) y + (2 - x^2) y' - (2 + x) y = 0
Linear Second Order ODE/(x^2 + x) y'' + (2 - x^2) y' - (2 + x) y = 0/Proof 1

== Theorem ==
== Proof ==
Note that:
and so:
Thus:
:$y_1 = \dfrac 1 x$
is a particular solution of $(1)$.
$(1)$ can be expressed as:
:$(2): \quad y + \dfrac {2 - x^2} {x^2 + x} y' - \dfrac {2 + x} {x^2 + x} y = 0$
which is in the form:
:$y + \map P x y' + \map Q x y = 0$
where:
:$\map P x = \dfrac {2 - x^2} {x^2 + x}$
:$\map Q x = \dfrac {2 + x} {x^2 + x}$
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:
Hence:
and so:
From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
:$y = C_1 e^x + \dfrac {C_2} x$
Category:Linear Second Order ODE/(x^2 + x) y + (2 - x^2) y' - (2 + x) y = 0
Linear Second Order ODE/(x^2 + x) y'' + (2 - x^2) y' - (2 + x) y = 0/Proof 2

== Theorem ==
== Proof ==
Note that:
:$\paren {x^2 + x} + \paren {2 - x^2} - \paren {2 + x} = 0$
so $\map {y_1} x$ such that $y_1 = {y_1}' = {y_1}$ satisfies $(1)$.
Hence:
:$y_1 = e^x$
is a particular solution of $(1)$.
$(1)$ can be expressed as:
:$(2): \quad y + \dfrac {2 - x^2} {x^2 + x} y' - \dfrac {2 + x} {x^2 + x} y = 0$
which is in the form:
:$y + \map P x y' + \map Q x y = 0$
where:
:$\map P x = \dfrac {2 - x^2} {x^2 + x}$
:$\map Q x = \dfrac {2 + x} {x^2 + x}$
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:
Hence:
and so:
From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
:$y = C_1 e^x + \dfrac {C_2} x$
Category:Linear Second Order ODE/(x^2 + x) y + (2 - x^2) y' - (2 + x) y = 0
Linear Second Order ODE/(x^2 + x) y'' + (2 - x^2) y' - (2 + x) y = x (x + 1)^2

== Theorem ==
The second order ODE:
:$(1): \quad \paren {x^2 + x} y + \paren {2 - x^2} y' - \paren {2 + x} y = x \paren {x + 1}^2$
has the general solution:
:$y = C_1 e^x + \dfrac {C_2} x - x - 1 - \dfrac {x^2} 3$
== Proof ==
$(1)$ can be manipulated into the form:
:$y + \dfrac {2 - x^2} {x^2 + x} y' - \dfrac {2 + x} {x^2 + x} y = x + 1$
It can be seen that this is a nonhomogeneous linear second order ODE in the form:
:$y + \map P x y' + \map Q x y = \map R x$
where:
:$\map P x = \dfrac {2 - x^2} {x^2 + x}$
:$\map Q x = -\dfrac {2 + x} {x^2 + x}$
:$\map R x = x + 1$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$\paren {x^2 + x} y + \paren {2 - x^2} y' - \paren {2 + x} y = 0$
From Linear Second Order ODE: $\paren {x^2 + x} y + \paren {2 - x^2} y' - \paren {2 + x} y = 0$, this has the general solution:
:$y_g = C_1 e^x + \dfrac {C_2} x$
It remains to find a particular solution $y_p$ to $(1)$.
Expressing $y_g$ in the form:
:$y_g = C_1 \map {y_1} x + C_2 \map {y_2} x$
we have:
By the Method of Variation of Parameters, we have that:
:$y_p = v_1 y_1 + v_2 y_2$
where:
where $\map W {y_1, y_2}$ is the Wronskian of $y_1$ and $y_2$.
We have that:
Hence:
It follows that:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 e^x + \dfrac {C_2} x - x - 1 - \dfrac {x^2} 3$
is the general solution to $(1)$.
== Sources ==
* : $\S 3.19$: Problem $4 \ \text{(b)}$
Category:Examples of Linear Second Order ODEs
Category:Examples of Method of Variation of Parameters
Linear Second Order ODE/(x^2 - 1) y'' - 2 x y' + 2 y = (x^2 - 1)^2

== Theorem ==
The second order ODE:
:$(1): \quad \paren {x^2 - 1} y - 2 x y' + 2 y = \paren {x^2 - 1}^2$
has the general solution:
:$y = C_1 x + C_2 \paren {x^2 + 1} + \dfrac {x^4} 6 - \dfrac {x^2} 2$
== Proof ==
$(1)$ can be manipulated into the form:
:$y - \dfrac {2 x} {x^2 - 1} y' + \dfrac 2 {x^2 - 1} y = x^2 - 1$
It can be seen that this is a nonhomogeneous linear second order ODE in the form:
:$y + \map P x y' + \map Q x y = \map R x$
where:
:$\map P x = -\dfrac {2 x} {x^2 - 1}$
:$\map Q x = \dfrac 2 {x^2 - 1}$
:$\map R x = x^2 - 1$
First we establish the solution of the corresponding homogeneous linear second order ODE:
:$\paren {x^2 - 1} y - 2 x y' + 2 y = 0$
From Second Order ODE: $\paren {x^2 - 1} y - 2 x y' + 2 y = 0$, this has the general solution:
:$y_g = C_1 x + C_2 \paren {x^2 + 1}$
It remains to find a particular solution $y_p$ to $(1)$.
Expressing $y_g$ in the form:
:$y_g = C_1 \map {y_1} x + C_2 \map {y_2} x$
we have:
By the Method of Variation of Parameters, we have that:
:$y_p = v_1 y_1 + v_2 y_2$
where:
where $\map W {y_1, y_2}$ is the Wronskian of $y_1$ and $y_2$.
We have that:
Hence:
It follows that:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 x + C_2 \paren {x^2 + 1} + \dfrac {x^4} 6 - \dfrac {x^2} 2$
is the general solution to $(1)$.
== Sources ==
* : $\S 3.19$: Problem $4 \ \text{(a)}$
Category:Examples of Linear Second Order ODEs
Category:Examples of Method of Variation of Parameters
Linear Second Order ODE/(x - 1) y'' - y' + y = 0

== Theorem ==
The second order ODE:
:$(1): \quad \paren {x - 1} y - x y' + y = 0$
has the general solution:
:$y = C_1 x + C_2 e^x$
== Proof ==
Note that:
and so by inspection:
:$y_1 = x$
is a particular solution of $(1)$.
$(1)$ can be expressed as:
:$(2): \quad y - \dfrac x {x - 1} y' + \dfrac 1 {x - 1} y = 0$
which is in the form:
:$y + \map P x y' + \map Q x y = 0$
where:
:$\map P x = -\dfrac x {x - 1}$
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:
Hence:
and so:
From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
:$y = C_1 x + C_2 e^x$
</onlyinclude>
== Sources ==
* : $\S 3.16$: Problem $7 \ \text{(a)}$
Category:Examples of Homogeneous LSOODEs
Linear Second Order ODE/2 x^2 y'' + 10 x y' + 8 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad 2 x^2 y' ' + 10 x y' + 8 y = 0$
has the general solution:
:$y = C_1 x^{-2} + C_2 x^{-2} \ln x$
== Proof ==
Let $(1)$ be rewritten as:
:$x^2 y' ' + 5 x y' + 4 y = 0$
It can be seen to be an instance of the Cauchy-Euler Equation:
:$x^2 y' ' + p x y' + q y = 0$
where:
:$p = 5$
:$q = 4$
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:
:$(2): \quad \dfrac {\d^2 y} {\d t^2} + 4 \dfrac {\d y} {\d t^2} + 4 y = 0$
From Linear Second Order ODE: $y' ' + 4 y' + 4 y = 0$, this has the general solution:
== Sources ==
* : $\S 3.17$: Problem $4 \ \text{(b)}$
Category:Examples of Cauchy-Euler Equation
Linear Second Order ODE/2 y'' + 2 y' + 3 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad 2 y + 2 y' + 3 y = 0$
has the general solution:
:$y = e^{-x/2} \paren {C_1 \cos \dfrac {\sqrt 5} 2 x + C_2 \sin \dfrac {\sqrt 5} 2 x}$
== Proof ==
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: 2 m^2 + 2 m + 3 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = -\dfrac 1 2 + \dfrac {\sqrt 5} 2 i$
:$m_2 = -\dfrac 1 2 - \dfrac {\sqrt 5} 2 i$
These are complex and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = e^{-x/2} \paren {C_1 \cos \dfrac {\sqrt 5} 2 x + C_2 \sin \dfrac {\sqrt 5} 2 x}$
== Sources ==
* : $\S 3.17$: Problem $1 \ \text{(g)}$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/2 y'' - 4 y' + 8 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad 2 y - 4 y + 8 y = 0$
has the general solution:
:$y = e^x \paren {C_1 \cos \sqrt 3 x + C_2 \sin \sqrt 3 x}$
== Proof ==
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Let $(1)$ be written in the form:
:$y - 2 y + 4 y = 0$
Its auxiliary equation is:
:$(2): \quad: m^2 - 2 m + 4 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = 1 + \sqrt 3 i$
:$m_2 = 1 - \sqrt 3 i$
These are complex and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = e^x \paren {C_1 \cos \sqrt 3 x + C_2 \sin \sqrt 3 x}$
== Sources ==
* : $\S 3.17$: Problem $1 \ \text{(d)}$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/x^2 y'' + 2 x y' - 12 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad x^2 y + 2 x y' - 12 y = 0$
has the general solution:
:$y = C_1 x^3 + C_2 x^{-4}$
== 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:
:$p = 2$
:$q = -12$
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:
:$(2): \quad \dfrac {\d^2 y} {\d t^2} + \dfrac {\d y} {\d t^2} - 12 y = 0$
From Linear Second Order ODE: $y + y' - 12 y = 0$, this has the general solution:
== Sources ==
* : $\S 3.17$: Problem $4 \ \text{(c)}$
Category:Examples of Cauchy-Euler Equation
Linear Second Order ODE/x^2 y'' + 2 x y' - 2 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad x^2 y + 2 x y' - 2 y = 0$
has the general solution:
:$y = C_1 x + \dfrac {C_2} {x^2}$
== Proof 1 ==
== Proof 2 ==
Category:Examples of Cauchy-Euler Equation
Category:Linear Second Order ODE/x^2 y + 2 x y' - 2 y = 0
Linear Second Order ODE/x^2 y'' + 2 x y' - 2 y = 0/Proof 1

== Theorem ==
== Proof ==
Note that:
and so by inspection:
:$y_1 = x$
is a particular solution of $(1)$.
$(1)$ can be expressed as:
:$(2): \quad y + \dfrac 2 x y' - \dfrac 2 {x^2} y = 0$
which is in the form:
:$y + \map P x y' + \map Q x y = 0$
where:
:$\map P x = \dfrac 2 x$
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:
Hence:
and so:
From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
:$y = C_1 x + k \paren {-\dfrac 1 {3 x^2} }$
where $k$ is arbitrary.
Setting $C_2 = -\dfrac k 3$:
:$y = C_1 x + \dfrac {C_2} {x^2}$
== Sources ==
* : $\S 3.16$: Problem $7 \ \text{(b)}$
Category:Linear Second Order ODE/x^2 y + 2 x y' - 2 y = 0
Linear Second Order ODE/x^2 y'' + 2 x y' - 2 y = 0/Proof 2

== Theorem ==
== 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: $y + y' - 2 y = 0$, this has the general solution:
Category:Linear Second Order ODE/x^2 y + 2 x y' - 2 y = 0
Linear Second Order ODE/x^2 y'' + 3 x y' + 10 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad x^2 y + 3 x y' + 10 y = 0$
has the general solution:
:$y = \dfrac 1 x \paren {C_1 \map \cos {\ln x^3} + C_2 \map \sin {\ln x^3} }$
== Proof ==
$(1)$ is an instance of the Cauchy-Euler Equation:
:$x^2 y + p x y' + q y = 0$
where:
:$p = 3$
:$q = 10$
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 $(1)$ can be expressed as:
:$(2): \quad \dfrac {\d^2 y} {\d t^2} + 2 \dfrac {\d y} {\d t^2} + 10 y = 0$
It can be seen that $(2)$ is a constant coefficient homogeneous linear second order ODE.
From Linear Second Order ODE: $y + 2 y' + 10 y = 0$, this has the general solution:
== Sources ==
* : $\S 3.17$: Problem $4 \ \text{(a)}$
Category:Examples of Cauchy-Euler Equation
Linear Second Order ODE/x^2 y'' + x y' - 4 y = 0

== Theorem ==
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 1 ==
== Proof 2 ==
Category:Examples of Cauchy-Euler Equation
Category:Linear Second Order ODE/x^2 y + x y' - 4 y = 0
Linear Second Order ODE/x^2 y'' + x y' - 4 y = 0/Proof 1

== Theorem ==
== Proof ==
Note that:
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:
Hence:
and so:
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:
:$y = C_1 x^2 + \dfrac {C_2} {x^2}$
== Sources ==
* : $\S 3.16$: Problem $4$
Category:Linear Second Order ODE/x^2 y + x y' - 4 y = 0
Linear Second Order ODE/x^2 y'' + x y' - 4 y = 0/Proof 2

== Theorem ==
== 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:
:$p = 1$
:$q = -4$
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:
:$(2): \quad \dfrac {\d^2 y} {\d t^2} - 4 y = 0$
From Linear Second Order ODE: $y - 4 y = 0$, this has the general solution:
Category:Linear Second Order ODE/x^2 y + x y' - 4 y = 0
Linear Second Order ODE/x^2 y'' + x y' - y = 0

== Theorem ==
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 1 ==
== Proof 2 ==
Category:Examples of Cauchy-Euler Equation
Category:Linear Second Order ODE/x^2 y + x y' - y = 0
Linear Second Order ODE/x^2 y'' + x y' - y = 0/Proof 1

== Theorem ==
== 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:
Hence:
and so:
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:
:$y = C_1 x + \dfrac {C_2} x$
== Sources ==
* : $\S 3.16$: The Use of a Known Solution to find Another: Example $1$
Category:Linear Second Order ODE/x^2 y + x y' - y = 0
Linear Second Order ODE/x^2 y'' + x y' - y = 0/Proof 2

== Theorem ==
== 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:
Category:Linear Second Order ODE/x^2 y + x y' - y = 0
Linear Second Order ODE/x^2 y'' - 2 x y' + 2 y = 0

== Theorem ==
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 1 ==
== Proof 2 ==
Category:Examples of Cauchy-Euler Equation
Category:Linear Second Order ODE/x^2 y - 2 x y' + 2 y = 0
Linear Second Order ODE/x^2 y'' - 2 x y' + 2 y = 0/Proof 1

== Theorem ==
== Proof ==
Consider the functions:
:$\map {y_1} x = x$
:$\map {y_2} x = x^2$
We have that:
Putting $x$ and $x^2$ into $(1)$ in turn:
Hence it can be seen that:
are particular solutions to $(1)$.
Calculating the Wronskian of $y_1$ and $y_2$:
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 $0$.
== Sources ==
* : $\S 3.15$: Problem $2$
Category:Linear Second Order ODE/x^2 y - 2 x y' + 2 y = 0
Linear Second Order ODE/x^2 y'' - 2 x y' + 2 y = 0/Proof 2

== Theorem ==
== 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:
:$\dfrac {\d^2 y} {\d t^2} + \paren {p - 1} \dfrac {\d y} {\d t^2} + q y = 0$
by making the substitution:
:$x = e^t$
Hence it can be expressed as:
:$(2): \quad \dfrac {\d^2 y} {\d t^2} - \dfrac {\d y} {\d t^2} + 2 y = 0$
From Linear Second Order ODE: $y - 3 y' + 2 y = 0$, this has the general solution:
Note that when $x = 0$, $e^t = x$ has no solution for $t$, and so is excluded from this solution.
Category:Linear Second Order ODE/x^2 y - 2 x y' + 2 y = 0
Linear Second Order ODE/x^2 y'' - 2 x y' + 2 y = 0/y(1) = 3, y'(1) = 5

== Theorem ==
The second order ODE:
:$(1): \quad x^2 y - 2 x y' + 2 y = 0$
with initial conditions:
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 $x$:
:$y' = C_1 + 2 C_2 x$
Thus for the initial conditions:
and:
Hence the result.
== Sources ==
* : $\S 3.15$: Problem $2$
Category:Linear Second Order ODE/x^2 y - 2 x y' + 2 y = 0
Linear Second Order ODE/y'' + 10 y' + 25 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad y' ' + 10 y' + 25 y = 0$
has the general solution:
:$y = C_1 e^{-5 x} + C_2 x e^{-5 x}$
== Proof ==
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 + 10 m + 25 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = m_2 = -5$
These are real and equal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = C_1 e^{-5 x} + C_2 x e^{-5 x}$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/y'' + 10 y' + 25 y = 14 exp -5 x

== Theorem ==
The second order ODE:
:$(1): \quad y + 10 y' + 25 y = 14 e^{-5 x}$
has the general solution:
:$y = C_1 \cos 2 x + C_2 \sin 2 x + \sin x$
== Proof ==
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form:
:$y + p y' + q y = \map R x$
where:
:$p = 10$
:$q = 25$
:$\map R x = 14 e^{-5 x}$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$(2): \quad y + 10 y' + 25 y = 0$
From Linear Second Order ODE: $y + 10 y' + 25 y = 0$, this has the general solution:
:$y_g = C_1 e^{-5 x} + C_2 x e^{-5 x}$
We have that:
:$\map R x = 14 e^{-5 x}$
and it is noted that $14 e^{-5 x}$ is a particular solution of $(2)$.
So from the Method of Undetermined Coefficients for the Exponential function:
:$y_p = A x^2 e^{-5 x}$
where:
:$A = \dfrac {14} 2$
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 e^{-5 x} + C_2 x e^{-5 x} + 7 x^2 e^{-5 x}$
== Sources ==
* : $\S 3.18$: Problem $1 \ \text{(c)}$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Undetermined Coefficients