nyuuzyou/wikis huggingface dataset:
Socrates is Mortal

== Theorem ==
:$(1): \quad$ All humans are mortal.
:$(2): \quad$ is human.
:$(3): \quad$ Therefore is mortal.
=== Variant ===
== Proof ==
Let $x$ be an object variable from the universe of rational beings.
Let $\map H x$ denote the propositional function $x$ is human.
Let $\map M x$ denote the propositional function $x$ is mortal.
Let $S$ be a proper name that denotes .
The argument can then be expressed as:
That is: 
: is mortal.
== Also presented as ==
== Historical Note ==
== Sources ==
* : Introduction
* : $1$ Introduction: Logic and Language: $1.2$: The Nature of Argument
* : $4$: Propositional Functions and Quantifiers: $4.2$: Proving Validity: Preliminary Quantification Rules
* : $1$. Pure Mathematics: Games
* 
* 
Category:Socrates is Mortal
Category:Logic
Category:Classic Problems
Socrates is Mortal/Variant

== Theorem ==
:$(1): \quad$ If is a man then is mortal.
:$(2): \quad$ is a man.
:$(3): \quad$ Therefore is mortal.
== Proof ==
Let $P$ denote the simple statement is a man..
Let $Q$ denote the simple statement is mortal..
The argument can then be expressed as:
That is: 
: is mortal.
== Also presented as ==
The subject of this syllogism varies.
For example, presents it as .
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.1$: Statements and connectives: Example $1.1$
Category:Socrates is Mortal
Solid Angle contained by Plane Angles is Less than Four Right Angles

== Theorem ==
== Proof ==
:
Let $A$ be a solid angle which is contained by the three plane angles $\angle BAC$, $\angle CAD$ and $\angle DAB$.
It is to be demonstrated that $\angle BAC + \angle CAD + \angle DAB$ is less than $4$ right angles.
Let $B$, $C$ and $D$ be arbitrary points on the straight lines $AB$, $AC$ and $AD$.
Let $BC$, $CD$ and $DB$ be joined.
We now have that the solid angle at $B$ is contained by the three plane angles $\angle CBA$, $\angle ABD$ and $\angle CBD$.
From :
:any two of these is greater than the other one.
Therefore $\angle CBA + \angle ABD > \angle CBD$.
For the same reason:
:$\angle BCA + \angle ACD > \angle BCD$
:$\angle CDA + \angle ADB > \angle CDB$
Therefore the sum of the six plane angles:
:$\angle CBA + \angle ABD + \angle BCA + \angle ACD + \angle CDA + \angle ADB$
are greater than the three plane angles:
:$\angle CBD + \angle BCD + \angle CDB$
But from :
:$\angle CBD + \angle BCD + \angle CDB$ equals $2$ right angles.
Therefore $\angle CBA + \angle ABD + \angle BCA + \angle ACD + \angle CDA + \angle ADB$ is greater than two right angles.
Also from :
:$\angle ABC + \angle ACD + \angle ADB$ equals $2$ right angles.
Therefore from :
:$\angle CBA + \angle ACB + \angle BAC + \angle ACD + \angle CDA + \angle CAD + \angle ADB + \angle DBA + \angle BAD$ equals $6$ right angles.
Of them, $\angle ABC + \angle BCA + \angle ACD + \angle CDA + \angle ADB + \angle DBA$ are greater than $2$ right angles.
Therefore the remaining three angles:
:$\angle BAC + \angle CAD + \angle DAB$
are less than $4$ right angles.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Solid Angles
Solution Space of Nonhomogeneous Linear Equation forms Affine Space

== Theorem ==
Let $\EE$ be a nonhomogeneous linear ordinary differential equation.
The solution to $\EE$ forms an affine space.
== Examples ==
== Proof ==
== Sources ==
* 
Category:Solution Space of Nonhomogeneous Linear Equation forms Affine Space
Category:Examples of Affine Spaces
Category:Linear ODEs
Solution Space of Nonhomogeneous Linear Equation forms Affine Space/Examples/Arbitrary Example 1

== Example of Solution Space of Nonhomogeneous Linear Equation forms Affine Space ==
Let $S$ be the solution set to the linear second order ordinary differential equation:
:$y' ' - y = 1$
Then:
:$S$ is an affine space
:the vector space $X$ is the solution set to the linear second order ordinary differential equation:
::$y' ' - y = 0$
:such that $S \times X \to S$ is the mapping which defines the affine space $S$.
== Sources ==
* 
Category:Solution Space of Nonhomogeneous Linear Equation forms Affine Space
Solution Space of Nonhomogeneous Linear Equation forms Affine Space/Mistake

== Source Work ==
:
:affine space
== Mistake ==
:For example, the solutions to the ODE $y' ' - y = 1$ has a solution set $S$ ...
== Correction ==
To talk about the "solution set to the solutions" of an ODE does not make a great deal of sense.
What it should be is probably something like this:
:For example, the ODE $y' ' - y = 1$ has a solution set $S$ ...
== Sources ==
* 
Category:Mistakes/The Concise Oxford Dictionary of Mathematics
Solution by Integrating Factor/Examples/y' + y = x^-1

== Theorem ==
Consider the linear first order ODE:
:$(1): \quad \dfrac {\d y} {\d x} + y = \dfrac 1 x$
with the initial condition $\tuple {1, 0}$.
This has the particular solution:
:$y = \ds e^{-x} \int_1^x \dfrac {e^\xi \rd \xi} \xi$
== Proof ==
This is a linear first order ODE with constant coefficents in the form:
:$\dfrac {\d y} {\d x} + a y = \map Q x$
where:
:$a = 1$
:$\map Q x = \dfrac 1 x$
with the initial condition $y = 0$ when $x = 1$.
Thus from Solution to Linear First Order ODE with Constant Coefficients with Initial Condition:
From Primitive of $\dfrac {e^x} x$ has no Solution in Elementary Functions, further work on this is not trivial.
== Also see ==
* Primitive of Exponential of a x over x
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 1$. The first order equation: $\S 1.2$ The integrating factor: Example $2$
Category:Examples of Solution by Integrating Factor
Solution by Integrating Factor/Examples/y' - 3y = sin x

== Theorem ==
The linear first order ODE:
:$\dfrac {\d y} {\d x} - 3 y = \sin x$
has the general solution:
:$y = \dfrac 1 {10} \paren {3 \sin x - \cos x} + C e^{3 x}$
== Proof 1 ==
== Proof 2 ==
Category:Examples of Solution by Integrating Factor
Category:Solution by Integrating Factor/Examples/y' - 3y
Solution by Integrating Factor/Examples/y' - 3y = sin x/Proof 1

== Theorem ==
== Proof ==
This is a linear first order ODE in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where:
:$\map P x y = -3 y$
:$\map Q x = \sin x$
Thus:
Thus from Solution by Integrating Factor:
Category:Solution by Integrating Factor/Examples/y' - 3y
Solution by Integrating Factor/Examples/y' - 3y = sin x/Proof 2

== Theorem ==
== Proof ==
This is a linear first order ODE with constant coefficents in the form:
:$\dfrac {\d y} {\d x} + a y = \map Q x$
where:
:$a = -3$
:$\map Q x = \sin x$
Thus from Solution to Linear First Order ODE with Constant Coefficients:
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 1$. The first order equation: $\S 1.2$ The integrating factor: Example $1$
Category:Solution by Integrating Factor/Examples/y' - 3y
Solution of Constant Coefficient Homogeneous LSOODE

== Theorem ==
Let:
:$(1): \quad y' ' + p y' + q y = 0$
be a constant coefficient homogeneous linear second order ODE.
Let $m_1$ and $m_2$ be the roots of the auxiliary equation $m^2 + p m + q = 0$.
Then $(1)$ has the general solution:
:$y = \begin{cases} C_1 e^{m_1 x} + C_2 e^{m_2 x} & : p^2 > 4 q \\ & \\ \paren {C_1 + C_2 x} e^{m_1 x} & : p^2 = 4 q \\ & \\ e^{a x} \paren {C_1 \sin b x + C_2 \cos b x} & : p^2 < 4 q \end{cases}$
where:
:$a + i b = m_1$
:$a - i b = m_2$
== Proof ==
=== Real Roots of Auxiliary Equation ===
=== Equal Real Roots of Auxiliary Equation ===
=== Complex Roots of Auxiliary Equation ===
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.1$ The reduced equation
* : $\S 18$: Basic Differential Equations and Solutions: $18.7$: Linear, homogeneous second order equation
* : $\S 3.17$: The Homogeneous Equation with Constant Coefficients
* : : $(3)$ Linear equations with constant coefficients of the form $a \dfrac {\d^2 y} {\d x^2} + b \dfrac {\d y} {\d x} + c y = 0$
* : : $(3)$ Linear equations with constant coefficients of the form $a \dfrac {\d^2 y} {\d x^2} + b \dfrac {\d y} {\d x} + c y = 0$
Category:Solution of Constant Coefficient Homogeneous LSOODE
Category:Constant Coefficient Homogeneous LSOODEs
Solution of Constant Coefficient Homogeneous LSOODE/Complex Roots of Auxiliary Equation

== Theorem ==
Let $p^2 < 4 q$.
Then $(1)$ has the general solution:
:$y = e^{a x} \paren {C_1 \cos b x + C_2 \sin b x}$
where:
:$m_1 = a + i b$
:$m_2 = a - i b$
== Proof ==
Consider the auxiliary equation of $(1)$:
:$(2): \quad m^2 + p m + q$
Let $p^2 < 4 q$.
From Solution to Quadratic Equation with Real Coefficients, $(2)$ has two complex roots:
As $p^2 < 4 q$ we have that:
:$\sqrt {q - \dfrac {p^2} 4} \ne 0$
and so:
:$m_1 \ne m_2$
Let:
where $a = -\dfrac p 2$ and $b = \sqrt {q - \dfrac {p^2} 4}$.
From Exponential Function is Solution of Constant Coefficient Homogeneous LSOODE iff Index is Root of Auxiliary Equation:
are both particular solutions to $(1)$.
We can manipulate $y_a$ and $y_b$ into the following forms:
and:
Hence:
Let:
We have that:
As $\cot b x$ is not zero for all $x$, $y_1$ and $y_2$ are linearly independent.
From Linear Combination of Solutions to Homogeneous Linear 2nd Order ODE:
:$y_1 = \dfrac {y_a + y_b} 2$
:$y_2 = \dfrac {y_b - y_b} 2$
are both particular solutions to $(1)$.
It follows from Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution that:
:$y = C_1 e^{a x} \cos b x + C_2 e^{a x} \sin b x$
or:
:$y = e^{a x} \paren {C_1 \cos b x + C_2 \sin b x}$
is the general solution to $(1)$.
== Sources ==
* : $\S 3.17$: The Homogeneous Equation with Constant Coefficients
Category:Solution of Constant Coefficient Homogeneous LSOODE
Solution of Constant Coefficient Homogeneous LSOODE/Equal Real Roots of Auxiliary Equation

== Theorem ==
Let $p^2 = 4 q$.
Then $(1)$ has the general solution:
:$y = C_1 e^{m_1 x} + C_2 x e^{m_1 x}$
== Proof ==
Consider the auxiliary equation of $(1)$:
:$(2): \quad m^2 + p m + q$
Let $p^2 = 4 q$.
From Solution to Quadratic Equation with Real Coefficients, $(2)$ has one (repeated) root, that is:
:$m_1 = m_2 = -\dfrac p 2$
From Exponential Function is Solution of Constant Coefficient Homogeneous LSOODE iff Index is Root of Auxiliary Equation:
:$y_1 = e^{m_1 x}$
is a particular solution to $(1)$.
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:
:$y = C_1 e^{m_1 x} + C_2 x e^{m_1 x}$
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.1$ The reduced equation
* : $\S 3.17$: The Homogeneous Equation with Constant Coefficients
Category:Solution of Constant Coefficient Homogeneous LSOODE
Solution of Constant Coefficient Homogeneous LSOODE/Real Roots of Auxiliary Equation

== Theorem ==
Let $p^2 > 4 q$.
Then $(1)$ has the general solution:
:$y = C_1 e^{m_1 x} + C_2 e^{m_2 x}$
== Proof ==
Consider the auxiliary equation of $(1)$:
:$(2): \quad m^2 + p m + q$
Let $p^2 > 4 q$.
From Solution to Quadratic Equation with Real Coefficients, $(2)$ has two real roots:
As $p^2 > 4 q$ we have that:
:$\sqrt {\dfrac {p^2} 4 - q} \ne 0$
and so:
:$m_1 \ne m_2$
From Exponential Function is Solution of Constant Coefficient Homogeneous LSOODE iff Index is Root of Auxiliary Equation:
are both particular solutions to $(1)$.
We also have that:
Thus $y_1$ and $y_2$ are linearly independent.
It follows from Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution that:
:$y = C_1 e^{m_1 x} + C_2 e^{m_2 x}$
is the general solution to $(1)$.
== Sources ==
* : $\S 3.17$: The Homogeneous Equation with Constant Coefficients
Category:Solution of Constant Coefficient Homogeneous LSOODE
Solution of Constant Coefficient LSOODE

== Proof Technique ==
Consider the linear second order ODE with constant coefficients:
:$(1): \quad \dfrac {\d^2 y} {\d x^2} + p \dfrac {\d y} {\d x} + q y = \map R x$
where $p$ and $q$ are constants and $\map R x$ is a function of $x$.
The general solution to $(1)$ can be found as follows.
;Find the roots $m_1$ and $m_2$ of the auxiliary equation $m^2 + p m + q = 0$
Hence the general solution $\map {y_g} x$ is found of the homogeneous linear second order ODE:
:$(2): \quad \dfrac {\d^2 y} {\d x^2} + p \dfrac {\d y} {\d x} + q y = 0$
using Solution of Constant Coefficient Homogeneous LSOODE.
;Select a trial solution
Using the Method of Undetermined Coefficients, identify a trial solution to $(1)$ by selecting a real function $\map f x$ of the same form as $\map R x$.
This can be done when $\map R x$ is in the form:
:a polynomial in $x$:
::$\ds \map R x = \sum_{j \mathop = 0}^n a_j x^j$
:of the form $K e^{a x}$
:of the form $A \cos a x + B \sin b x$
or a sum or product of these.
;Solve the trial solution
Differentiate $f$ twice $x$.
Then substitute for $y$, $y'$ and $y$ in $(1)$ to obtain a system of simultaneous equations from which the so far undetermined coefficients of $f$ can be evaluated.
Hence a particular solution $y_p$ to $(1)$ is obtained.
;Other cases
In the cases when $\map R x$ is not of one of the above forms, $(1)$ can be split into its factors as follows.
:Express $(1)$ in the form $\paren {D^2 + p D + q} y = \map R x$
:Express it further in the form $\paren {D + m_1} \paren {D + m_2} y = \map R x$
:Solve the resulting linear first Order ODE with constant coefficients.
The general solution to $(1)$ is then:
:$\map y x = \map {y_g} x + \map {y_p} x$
== Sources ==
* : $\S 18$: Basic Differential Equations and Solutions: $18.8$: Linear, nonhomogeneous second order equation
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.8$ Recapitulation
* : : $(4)$ Linear equations with constant coefficients of the form $a \dfrac {\d^2 y} {\d x^2} + b \dfrac {\d y} {\d x} + c y = \map f x$
* : : $(4)$ Linear equations with constant coefficients of the form $a \dfrac {\d^2 y} {\d x^2} + b \dfrac {\d y} {\d x} + c y = \map f x$
Category:Constant Coefficient LSOODEs
Solution of Constant Coefficient Linear nth Order ODE

== Proof Technique ==
Consider the linear second order ODE with constant coefficients:
:$(1): \quad \ds \sum_{k \mathop = 0}^n a_k \dfrac {\d^k y} {d x^k} = \map R x$
where $a_k$ is a constant for $0 \le k \le n$ and $\map R x$ is a function of $x$.
The general solution to $(1)$ can be found as follows.
;Find the roots $m_1, m_2, \ldots, m_n$ of the auxiliary equation:
:$(2): \quad \ds \sum_{k \mathop = 0}^n a_k m^k = 0$.
If $(2)$ has distinct roots, then the general solution $\map {y_g} x$ is of the form:
:$\ds y_g = \sum_{k \mathop = 0}^n A_k e^{m_k x}$
If there are repeated roots of $(2)$, further needs to be done.
Let $m_j$ be a repeated root of $(2)$ with multiplicity $r$.
Then the of $(2)$ can be written:
:$\map P m \paren {m - m_j}^r$
where $\map P m$ is a polynomial which does not contain the factor $m - m_j$.
Then the $r$ instances of $m_j$ give rise to the solutions:
:$(4): \quad y = A_{j_0} e^{m_j x} + A_{j_1} x e^{m_j x} + \dotsb + A_{j_{r - 1} } x^{r - 1} e^{m_j x}$
== Proof ==
Let the reduced equation of $(1)$ be written:
:$(3): \quad \paren {a_n D^n + a_{n - 1} D^{n - 1} + \dotsb + a_1 D + a_0} y = 0$
By factoring the we get:
:$\map P D \paren {D - m_j}^r y = 0$
It remains to be shown that $(4)$ is a solution to $(3)$.
Each of the terms is of the form $p e^{m_j x} u$, where $u$ is a power of $x$ and $p$ is a constant.
So:
and so on until:
because $D^r u = 0$ when $u = 1, x, x^2, \ldots, x^{r - 1}$.
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 3$. Equations of higher order and systems of first order equations: $\S 3.1$ The $n$th order equation
Category:Linear ODEs
Solution of Equation/Examples

== Examples of Solution Sets ==
=== Solution Set of $x^2 = 2$ in $\R$ ===
=== Solution Set of $x^2 = 2$ in $\Q$ ===
=== Solution to $x^2 - 2 x - 3 = 0$ ===
=== Solution to $x^2 - x - 6 = 0$ ===
=== Solution to $x^3 + x^2 - 3 x - 3 = 0$ ===
=== Solution to $x^2 + 49 = 0$ ===
Category:Examples of Solution Sets
Solution of Equation/Examples/x^2 + 49

== Example of Solution Set ==
Consider the equation in algebra:
:$x^2 + 49 = 0$
where the domain of $x$ is implicitly taken to be the set of complex numbers $\C$.
Then the solutions to $x^2 + 49 = 0$ are $7 i$ and $-7 i$.
== Proof ==
We have that:
and the result follows.
== Sources ==
* 
* 
Category:Examples of Solution Sets
Solution of Equation/Examples/x^2 - 2x - 3

== Example of Solution Set ==
Consider the equation in algebra:
:$x^2 - 2 x - 3 = 0$
where the domain of $x$ is implicitly taken to be the set of real numbers $\R$.
Then $3$ is a solution to $x^2 - 2 x - 3 = 0$.
== Proof ==
We have that:
and the result follows by definition of solution.
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation
Category:Examples of Solution Sets
Solution of Equation/Examples/x^2 - x - 6

== Example of Solution Set ==
Consider the equation in algebra:
:$x^2 - x - 6 = 0$
where the domain of $x$ is implicitly taken to be the set of real numbers $\R$.
Then the solutions to $x^2 - x - 6 = 0$ are $3$ and $-2$.
== Proof ==
We have that:
and the result follows.
== Sources ==
* 
* 
Category:Examples of Solution Sets
Solution of Equation/Examples/x^2 = 2 in Rationals

== Example of Solution Set ==
Let $x$ denote a variable whose domain is the set of real numbers $\Q$.
Let $\map P x$ be the propositional function defined as:
:$\map P x := x^2 - 2$
Then the solution set of $\map P x$ is the empty set $\O$.
== Proof ==
From Difference of Two Squares, we have:
:$x^2 - 2 = \paren {x - \sqrt 2} \paren {x + \sqrt 2}$
from which it follows that:
:$x = \sqrt 2$
or:
:$x = -\sqrt 2$
As both $\sqrt 2 \notin \Q$ and $-\sqrt 2 \notin \Q$ the result follows.
== Sources ==
* : $\S 1$: Some mathematical language: Variables and quantifiers
Category:Examples of Solution Sets
Solution of Equation/Examples/x^2 = 2 in Reals

== Example of Solution Set ==
Let $x$ denote a variable whose domain is the set of real numbers $\R$.
Let $\map P x$ be the propositional function defined as:
:$\map P x := x^2 - 2$
Then the solution set of $\map P x$ is $\set {\sqrt 2, -\sqrt 2}$.
== Proof ==
From Difference of Two Squares, we have:
:$x^2 - 2 = \paren {x - \sqrt 2} \paren {x + \sqrt 2}$
from which it follows that:
:$x = \sqrt 2$
or:
:$x = -\sqrt 2$
As both $\sqrt 2 \in \R$ and $-\sqrt 2 \in \R$ the result follows.
== Sources ==
* : $\S 1$: Some mathematical language: Variables and quantifiers
Category:Examples of Solution Sets
Solution of Equation/Examples/x^3 + x^2 - 3x - 3

== Example of Solution Set ==
Consider the equation in algebra:
:$x^3 + x^2 - 3 x - 3 = 0$
where the domain of $x$ is implicitly taken to be the set of real numbers $\R$.
Then the solutions to $x^3 + x^2 - 3 x - 3 = 0$ are $\sqrt 3$, $-\sqrt 3$ and $-1$.
== Proof ==
We have that:
and the result follows.
== Sources ==
* 
* 
Category:Examples of Solution Sets
Solution of Equation using Integral Transform

== Proof Technique ==
Suppose it is necessary to solve a function $\map f x$ of some variable $x$.
Suppose also that $\map f x$ has to satisfy certain conditions expressed as equations which involve $\map f x$.
It may very well be the case that these equations are difficult to solve in the domain of $f$.
By application of an integral operator $T$, it is often possible to transform the equations specifying $\map f x$ into an equivalent set of equations in $\map F p$ which are easier to solve.
If a solution for $\map F p$ can be found, by use of the appropriate inversion theorem, $\map F p$ can then be converted back into the required solution for $\map f x$.
Problem: Equations involving $\map f x$		$\xrightarrow {\text {Integral operator } T}$ 		Equations involving $\map F p$		 	
 		 		$\Big \downarrow$ 		Solve these equations	
Solution for $\map f x$ 		$\xleftarrow {\text {Inverse integral operator } T^{-1} }$		Solution for $\map F p$ 		 	
The usual problem space for which integral transforms are most fruitfully used is the field of differential equations.
An integral transform will in this circumstance be used to convert a differential equation with its boundary conditions or initial conditions into a series of algebraic equations, which are in general more easily solved.
The appropriate use of the inverse integral operator converts the solution of this algebraic equation system back into that of the problem domain.
It is necessary to select of an appropriate integral transform which allows this to be accomplished.
== Sources ==
* : $\S 1.2$. The Usefulness of Integral Transforms
Category:Integral Transforms
Category:Proof Techniques
Solution of Linear 2nd Order ODE Tangent to X-Axis

== Theorem ==
Let $\map {y_p} x$ be a particular solution to the homogeneous linear second order ODE:
:$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
on a closed interval $\closedint a b$.
Let there exist $\xi \in \closedint a b$ such that the curve in the cartesian plane described by $y = \map {y_p} x$ is tangent to the $x$-axis at $\xi$.
Then $\map {y_p} x$ is the zero constant function:
:$\forall x \in \closedint a b: \map {y_p} x = 0$
== Proof ==
 $y_p$ is not the zero constant function.
From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another, there exists another particular solution to $(1)$ such that $y_1$ and $y_2$ are linearly independent.
At the point $\xi$:
:$\map {y_p} \xi = 0$
:$\map { {y_p}'} \xi = 0$
Taking the Wronskian of $y_p$ and $y_2$:
:$\map W {y_p, y_2} = y_p {y_2}' - {y_p}' y_2$
But at $\xi$ this works out as zero.
It follows from Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent that $y_p$ and $y_2$ cannot be linearly independent after all.
From this contradiction, $y_p$ must the zero constant function.
== Sources ==
* : $\S 3.14$: Problem $6$
Category:Linear Second Order ODEs
Solution of Linear Congruence/Examples

== Examples of Solutions of Linear Congruences ==
=== Solution to $5 x = 4 \pmod 3$ ===
=== Solution to $7 x = 6 \pmod 5$ ===
=== Solution to $9 x = 8 \pmod 7$ ===
=== Solution to $6 x = 5 \pmod 4$ ===
</onlyinclude>
=== Solution to $10 x = 8 \pmod 6$ ===
</onlyinclude>
=== Solution to $12 x = 9 \pmod 6$ ===
</onlyinclude>
=== Solution to $325 n = 11 \pmod 3$ ===
Category:Solution of Linear Congruence
Solution of Linear Congruence/Examples/10 x = 8 mod 6

== Example of Solution of Linear Congruence ==
Let $10 x = 8 \pmod 6$.
Then:
:$x = 4 + 7 t$
where $t \in \Z$.
== Proof ==
We have that:
:$8 \equiv 2 \pmod 6$
and so:
That should simplify the arithmetic.
Then:
From Solution of Linear Diophantine Equation, the general solution to $(1)$ is:
:$(2): \quad \forall t \in \Z: x = x_0 + 3 t, k = k_0 + 5 t$
where $x_0, k_0$ can be found as follows.
Using the Euclidean Algorithm:
Thus we have that:
:$\gcd \set {10, -6} = 2$
which is (trivially) a divisor of $2$.
So, from Solution of Linear Diophantine Equation, a solution exists.
Next we find a single solution to $10 x - 6 k = 2$.
Again with the Euclidean Algorithm:
and so:
is a solution.
Thus from $(2)$:
:$x = 2 + 3 t$
== Sources ==
* : $\text {4-1}$ Basic Properties of Congruences: Exercise $2 \ \text{(b)}$
Category:Examples of Modulo Arithmetic
Solution of Linear Congruence/Examples/12 x = 9 mod 6

== Example of Solution of Linear Congruence ==
Let $12 x = 9 \pmod 6$.
Then $x$ has no solution in $\Z$.
== Proof ==
We have that:
Then we have that:
:$\gcd \set {12, -6} = 6$
which is not a divisor of $9$.
So, from Solution of Linear Diophantine Equation, no solution exists.
== Sources ==
* : $\text {4-1}$ Basic Properties of Congruences: Exercise $2 \ \text{(c)}$
Category:Examples of Modulo Arithmetic