nyuuzyou/wikis huggingface dataset:
Solution to Exact Differential Equation

== Theorem ==
The first order ordinary differential equation‎:
:$F = \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
is an exact differential equation‎ :
:$\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$
The general solution of such an equation‎ is:
:$\map f {x, y} = C$
where:
:$\dfrac {\partial f} {\partial x} = M$
:$\dfrac {\partial f} {\partial y} = N$
== Proof ==
=== Necessary Condition ===
Let $F$ be exact.
Then by definition there exists a function whose second partial derivatives of $f$ exist and are continuous:
:$\map f {x, y}$
where:
:$\dfrac {\partial f} {\partial x} = M$
:$\dfrac {\partial f} {\partial y} = N$
Differentiating $M$ and $N$ partially $y$ and $x$ respectively:
:$\dfrac {\partial M} {\partial y} = \dfrac {\partial^2 f} {\partial x \partial y}$
:$\dfrac {\partial N} {\partial x} = \dfrac {\partial^2 f} {\partial y \partial x}$
From Mixed Second Partial Derivatives are Equal:
:$\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$
=== Sufficient Condition ===
Let $F$ be such that:
:$\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$
Take the equation:
:$\dfrac {\partial f} {\partial x} = M$
and integrate it $x$:
:$(1): \quad f = \ds \int M \rd x + \map g y$
Note that the arbitrary constant here is an arbitrary function of $y$, since when differentiating partially $x$ it disappears.
So, now we need to find such a $\map g y$ so as to make $f$ as defined in $(1)$ satisfy $\dfrac {\partial f} {\partial y} = N$.
Differentiating $(1)$ $y$ and equating it to $N$:
:$\ds \dfrac {\partial} {\partial y} \int M \rd x + \map {g'} y = N$
So:
:$\ds \map {g'} y = N - \dfrac {\partial} {\partial y} \int M \rd x$
which can be integrated $y$ thus:
:$\ds \map g y = \int \paren {N - \frac {\partial} {\partial y} \int M \rd x} \rd y$
which works as long as the integrand is a function of $y$ only.
This will happen if its derivative $x$ is equal to zero.
We need to make sure of that, so we try it out: 
This is our initial condition.
== The Solution ==
Then $F$ can be written in the form:
:$\dfrac {\partial f} {\partial x} \rd x + \dfrac {\partial f} {\partial y} \rd y = 0$
and the solution follows.
== Also see ==
== Sources ==
* : $\S 18$: Basic Differential Equations and Solutions: $18.4$: Exact equation
* : $\S 2.8$: Exact Equations
* : : $(1)$ Exact equations
* : : $(1)$ Exact equations
Category:Exact Differential Equations
Solution to First Order Initial Value Problem

== Theorem ==
Let $\map y x$ be a solution to the first order ordinary differential equation:
:$\dfrac {\d y} {\d x} = \map f {x, y}$
which is subject to an initial condition: $\tuple {a, b}$.
Then this problem is equivalent to the integral equation:
:$\ds y = b + \int_a^x \map f {t, \map y t} \rd t$
== Proof ==
From Solution to First Order ODE, the general solution of:
:$\dfrac {\d y} {\d x} = \map f {x, y}$
is:
:$\ds y = \int \map f {x, \map y x} \rd x + C$
When $x = a$, we have $y = b$.
Thus:
:$\ds b = \valueat {\int \map f {x, \map y x} \rd x + C} a$
which gives:
:$\ds C = b - \valueat {\int \map f {x, \map y x} \rd x} a$
and so:
$\ds y = b + \int \map f {x, \map y x} \rd x - \valueat {\int \map f {x, \map y x} \rd x} a$
whence the result, by the Fundamental Theorem of Calculus.
Category:First Order ODEs
Solution to First Order ODE

== Theorem ==
Let:
:$\Phi = \dfrac {\d y} {\d x} = \map f {x, y}$
be a first order ordinary differential equation.
Then $\Phi$ has a general solution which can be expressed in terms of an indefinite integral of $\map f x$:
:$\ds y = \int \map f {x, y} \rd x + C$
where $C$ is an arbitrary constant.
== Proof ==
Integrating both sides with respect to $x$:
The validity of this follows from Picard's Existence Theorem.
Category:First Order ODEs
Solution to Gambler's Ruin

== Theorem ==
Let a gambler $G$ with an initial capital of $C$ units play a sequence of games in which:
:$G$ gains $1$ unit of capital with probability $p$
:$G$ loses $1$ unit of capital with probability $q = 1 - p$.
The game end either when:
:$G$ is ruined he loses all $C$ units
:$G$ wins he attains a total fortune of $N$ units, where $N > C$.
Then the probability that $G$ is ruined is given by:
:$\map \Pr {\text {ruin} } = \begin {cases} \dfrac {N - C} N & : p = \dfrac 1 2 \\ \\ q^C \dfrac {p^{N - C} - q^{N - C} } {p^N - q^N} & : \text {otherwise} \end {cases}$
== Proof ==
== Sources ==
* 
Category:Gambler's Ruin
Solution to Hawk-Dove

== Solution to Hawk-Dove ==
== Proof ==
From the payoff table:
Each player gains most by acting as a hawk while the other player acts as a dove.
Each player loses most when both act as hawks.
So each animal prefers to be a hawk if its opponent prefers to be a dove.
Thus there are two Nash equilibria:
:$\tuple {\text{Hawk}, \text{Dove} }$
:$\tuple {\text{Dove}, \text{Hawk} }$
== Sources ==
* : $\text I$ Strategic Games: Chapter $2$ Nash Equilibrium: $2.3$: Examples: Example $16.3$
Category:Examples of Two-Person Games
Solution to Homogeneous Differential Equation

== Theorem ==
Let:
:$\map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
be a homogeneous differential equation‎.
It can be solved by making the substitution $z = \dfrac y x$.
Its solution is:
:$\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
where:
:$\map f {x, y} = -\dfrac {\map M {x, y} } {\map N {x, y} }$
== Proof ==
From the original equation‎, we see:
:$\dfrac {\d y} {\d x} = \map f {x, y} = -\dfrac {\map M {x, y} } {\map N {x, y} }$
From Quotient of Homogeneous Functions‎ it follows that $\map f {x, y}$ is homogeneous of degree zero.
Thus:
:$\map f {t x, t y} = t^0 \map f {x, y} = \map f {x, y}$
Set $t = \dfrac 1 x$ in this equation‎:
where $z = \dfrac y x$.
Then:
This is seen to be a separable differential equation.
On performing the required integrations and simplifying as necessary, the final step is to substitute $\dfrac y x$ back for $z$.
== Historical Note ==
== Sources ==
* : $\S 18$: Basic Differential Equations and Solutions: $18.5$: Homogeneous equation
* : $\S 2.7$: Homogeneous Equations
* : : $(3)$ Homogeneous equations
* : : $(3)$ Homogeneous equations
* : $\S 2.2$
* 
Category:Homogeneous Differential Equations
Solution to Hypergeometric Differential Equation

== Theorem ==
The hypergeometric series:
defines a solution to the hypergeometric differential equation:
:$x \paren {1 - x} \dfrac {\d^2 y} {\d x^2} + \paren {c - \paren {a + b + 1} x} \dfrac {\d y} {\d x} - a b y = 0$
== Proof ==
Let:
Then:
And:
Therefore:
== Also see ==
* Definition:Gaussian Hypergeometric Function
== Sources ==
* 
* 
Category:Gaussian Hypergeometric Function
Category:Hypergeometric Differential Equations
Category:Hypergeometric Series
Solution to Iterated Prisoner's Dilemma

== Solution to Iterated Prisoner's Dilemma ==
The optimal strategy for the iterated prisoner's dilemma for either player is:
:$(1): \quad$ Adopt the cooperating strategy on the first round.
:$(2): \quad$ On subsequent rounds, copy the previous move of the other player.
This is known as a tit-for-tat strategy.
== Proof ==
== Sources ==
* 
Category:Iterated Prisoner's Dilemma
Category:Tit-For-Tat Strategies
Solution to Legendre's Differential Equation

== Theorem ==
The solution of Legendre's differential equation:
can be obtained by Power Series Solution Method.
== Proof ==
Let:
:$\ds y = \sum_{n \mathop = 0}^\infty a_n x^{k - n}$
such that:
:$a_0 \ne 0$
Differentating $x$:
:$\ds \dot y = \sum_{n \mathop = 0}^\infty a_n \paren {k - n} x^{k - n - 1}$
:$\ds \ddot y = \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2}$
Substituting in the original equation:
The summations are dependent upon $n$ and not $x$.
Therefore it is a valid operation to multiply the $x$'s into the summations, thus:
Increasing the summation variable $n$ to $n + 2$:
Taking the first 2 terms of the second summation out:
Equating each term to $0$:
Take equation $(1)$:
:$a_0 x^k \paren {p \paren {p + 1} - k \paren {k + 1} } = 0$
It is assumed that $a_0 \ne 0$ and $x^k$ can never be zero for any value of $k$.
Thus:
Take equation $(2)$:
:$a_1 x^{k - 1} \paren {p \paren {p + 1} - k \paren {k - 1} } = 0$
As before, it is assumed that $x^{k - 1}$ can never be zero for any value of $k$.
Thus:
Take equation $(3)$:
:$\ds \sum_{n \mathop = 2}^\infty x^{k - n} \paren {a_{n - 2} \paren {k - n - 2} \paren {k - n + 1} + a_n \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} } } = 0$
As before, it is assumed that $x^{k - n}$ can never be zero for any value of $k$.
Thus:
Since Legendre's differential equation is a second order ODE, it has two independent solutions.
=== Solution 1 (for $k = p$) ===
Thus:
Similarly:
Also:
:$a_3 = a_5 = a_7 = \dotsb = a_{2 n + 1} = a_1 = 0$
Substituting for $a_n$:
=== Solution 2 (for $k = p - 1$) ===
Thus:
Similarly:
Also:
:$a_3 = a_5 = a_7 = \dotsb = a_{2 n + 1} = a_1 = 0$
Substituting for $a_n$:
In summary, the two particular solutions are:
and:
== Also see ==
The solutions of Legendre's differential equation are known as Legendre polynomials, which are functions of the parameter $p$.
== Sources ==
* : $\S 18$: Basic Differential Equations and Solutions: $18.12$: Legendre's equation
* : $\S 19$: Basic Differential Equations and Solutions: $19.12$: Legendre's equation
Category:Legendre's Differential Equation
Category:Second Order ODEs
Solution to Linear First Order ODE with Constant Coefficients

== Theorem ==
A linear first order ODE with constant coefficients in the form:
:$(1): \quad \dfrac {\d y} {\d x} + a y = \map Q x$
has the general solution:
:$\ds y = e^{-a x} \paren {\int e^{a x} \map Q x \rd x + C}$
=== With Initial Condition ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 1$. The first order equation: $\S 1.2$ The integrating factor
Category:Linear First Order ODEs with Constant Coefficients
Category:Solution to Linear First Order ODE with Constant Coefficients
Solution to Linear First Order ODE with Constant Coefficients/Proof 1

== Theorem ==
== Proof ==
From the Product Rule for Derivatives:
Hence, multiplying $(1)$ all through by $e^{\int a \rd x}$:
:$\map {\dfrac \d {\d x} } {e^{a x} y} = e^{a x} \map Q x$
Integrating $x$ now gives:
:$\ds e^{a x} y = \int e^{a x} \map Q x \rd x + C$
whence we get the result by dividing by $e^{a x}$.
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 1$. The first order equation: $\S 1.2$ The integrating factor: $(7)$
Category:Solution to Linear First Order ODE with Constant Coefficients
Solution to Linear First Order ODE with Constant Coefficients/Proof 2

== Theorem ==
== Proof ==
This is a specific instance of Solution to Linear First Order Ordinary Differential Equation:
In this instance, we have:
:$\map P x = a$
Hence:
:$\ds \int \map P x = a x$
and the result follows.
Category:Solution to Linear First Order ODE with Constant Coefficients
Solution to Linear First Order ODE with Constant Coefficients/With Initial Condition

== Theorem ==
Consider the linear first order ODE with constant coefficients in the form:
:$(1): \quad \dfrac {\d y} {\d x} + a y = \map Q x$
with initial condition $\tuple {x_0, y_0}$
Then $(1)$ has the particular solution:
:$\ds y = e^{-a x} \int_{x_0}^x e^{a \xi} \map Q \xi \rd \xi + y_0 e^{a \paren {x - x_0} }$
== Proof ==
From Solution to Linear First Order ODE with Constant Coefficients, the general solution to $(1)$ is:
:$(2): \quad \ds y = e^{-a x} \int e^{a x} \map Q x \rd x + C e^{-a x}$
Let $y = y_0$ when $x = x_0$.
We have:
:$(3): \ds \quad y_0 = e^{-a x_0} \int e^{a x_0} \map Q {x_0} \rd x_0 + C e^{-a x_0}$
Thus:
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 1$. The first order equation: $\S 1.2$ The integrating factor: $(8)$
Category:Linear First Order ODEs with Constant Coefficients
Solution to Linear First Order Ordinary Differential Equation

== Theorem ==
A linear first order ordinary differential equation in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
has the general solution:
:$\ds y = e^{-\int P \rd x} \paren {\int Q e^{\int P \rd x} \rd x + C}$
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
== Also see ==
* Solution to Linear First Order ODE with Constant Coefficients, a specific instance
== Sources ==
* : $\S 18$: Basic Differential Equations and Solutions: $18.2$: Linear first order equation
* : : $(5)$ Linear equations
* : : $(5)$ Linear equations
Category:Solution to Linear First Order Ordinary Differential Equation
Category:Linear First Order ODEs
Solution to Linear First Order Ordinary Differential Equation/Also presented as

== Solution to Linear First Order Ordinary Differential Equation: Also presented as ==
The Solution to Linear First Order Ordinary Differential Equation is also presented as:
:$\ds y e^{\int P \rd x} = \int Q e^{\int P \rd x} \rd x + C$
== Sources ==
* : $\S 18$: Basic Differential Equations and Solutions: $18.2$: Linear first order equation
Category:Solution to Linear First Order Ordinary Differential Equation
Solution to Linear First Order Ordinary Differential Equation/Proof 1

== Theorem ==
== Proof ==
Consider the first order ordinary differential equation:
:$\map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
We can put our equation:
:$(1): \quad \dfrac {\d y} {\d x} + \map P x y = \map Q x$
into this format by identifying:
:$\map M {x, y} \equiv \map P x y - \map Q x, \map N {x, y} \equiv 1$
We see that:
:$\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x} = \map P x$
and hence:
:$\map P x = \dfrac {\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x} } N$
is a function of $x$ only.
It immediately follows from Integrating Factor for First Order ODE that:
:$e^{\int \map P x \rd x}$
is an integrating factor for $(1)$.
So, multiplying $(1)$ by this factor:
:$e^{\int \map P x \rd x} \dfrac {\d y} {\d x} + e^{\int \map P x \rd x} \map P x y = e^{\int \map P x \rd x} \map Q x$
The result follows by an application of Solution to Exact Differential Equation.
== Sources ==
* : $\S 2.10$: Linear Equations: Problem $1$
Category:Solution to Linear First Order Ordinary Differential Equation
Solution to Linear First Order Ordinary Differential Equation/Proof 2

== Theorem ==
== Proof ==
From the Product Rule for Derivatives:
Hence, multiplying $(1)$ all through by $e^{\int \map P x \rd x}$:
:$\dfrac \d {\d x} \paren {e^{\int \map P x \rd x} y} = \map Q x e^{\int \map P x \rd x}$
Integrating $x$ now gives:
:$\ds e^{\int \map P x \rd x} y = \int \map Q x e^{\int \map P x \rd x} \rd x + C$
whence we get the result by dividing by $e^{\int \map P x \rd x}$.
== Sources ==
* : $\S 2.10$
Category:Solution to Linear First Order Ordinary Differential Equation
Solution to Linear First Order Ordinary Differential Equation/Solution by Integrating Factor

== Proof Technique ==
The technique to solve a linear first order ordinary differential equation in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
It immediately follows from Integrating Factor for First Order ODE that:
:$e^{\int \map P x \rd x}$
is an integrating factor for $(1)$.
Multiplying it by:
:$e^{\int \map P x \rd x}$
to reduce it to a form:
:$\dfrac {\d y} {\d x} e^{\int \map P x \rd x} y = e^{\int \map P x \rd x} \map Q x$
is known as Solution by Integrating Factor.
It is remembered by the procedure:
:Multiply by $e^{\int \map P x \rd x}$ and integrate.
== Examples ==
== Sources ==
* : $\S 2.10$
Category:Linear First Order ODEs
Category:Proof Techniques
Solution to Pell's Equation

== Theorem ==
Recall Pell's equation:
Let the continued fraction of $\sqrt n$ have a cycle whose length is $s$:
:$\sqrt n = \sqbrk {a_1 \sequence {a_2, a_3, \ldots, a_{s + 1} } }$
Let $a_n = \dfrac {p_n} {q_n}$ be a convergent of $\sqrt n$.
Then:
:${p_{r s} }^2 - n {q_{r s} }^2 = \paren {-1}^{r s}$ for $r = 1, 2, 3, \ldots$
and all solutions of:
:$x^2 - n y^2 = \pm 1$
are given in this way.
== Proof ==
First note that if $x = p, y = q$ is a positive solution of $x^2 - n y^2 = 1$ then $\dfrac p q$ is a convergent of $\sqrt n$.
The continued fraction of $\sqrt n$ is periodic from Continued Fraction Expansion of Irrational Square Root and of the form:
:$\sqbrk {a \sequence {b_1, b_2, \ldots, b_{m - 1}, b_m, b_{m - 1}, \ldots, b_2, b_1, 2 a} }$
or 
:$\sqbrk {a \sequence {b_1, b_2, \ldots, b_{m - 1}, b_m, b_m, b_{m - 1}, \ldots, b_2, b_1, 2 a} }$
For each $r \ge 1$ we can write $\sqrt n$ as the (non-simple) finite continued fraction:
:$\sqrt n = \sqbrk {a \sequence {b_1, b_2, \ldots, b_2, b_1, 2 a, b_1, b_2, \ldots, b_2, b_1, x} }$
which has a total of $r s + 1$ partial denominators.
The last element $x$ is of course not an integer.
What we do have, though, is:
The final three convergents in the above FCF are:
:$\dfrac {p_{r s - 1} } {q_{r s - 1} }, \quad \dfrac {p_{r s} } {q_{r s} }, \quad \dfrac {x p_{r s} + p_{r s - 1} } {x q_{r s} + q_{r s - 1} }$
The last one of these equals $\sqrt n$ itself.
So:
:$\sqrt n \paren {x q_{r s} + q_{r s - 1} } = \paren {x p_{r s} + p_{r s - 1} }$
Substituting $a + \sqrt n$ for $x$, we get:
:$\sqrt n \paren {\paren {a + \sqrt n} q_{r s} + q_{r s - 1} } = \paren {\paren {a + \sqrt n} p_{r s} + p_{r s - 1} }$
This simplifies to:
:$\sqrt n \paren {a q_{r s} + q_{r s - 1} - p_{r s} } = a p_{r s} + p_{r s - 1} - n q_{r s}$
The of this is an integer while the is $\sqrt n$ times an integer.
Since $\sqrt n$ is irrational, the only way that can happen is if both sides equal zero.
This gives us:
Multiplying $(1)$ by $p_{r s}$, $(2)$ by $q_{r s}$ and then subtracting:
:$p_{r s}^2 - n q_{r s}^2 = p_{r s} q_{r s - 1} - p_{r s - 1} q_{r s}$
By Difference between Adjacent Convergents of Simple Continued Fraction, the of this is $\paren {-1}^{r s}$.
When the cycle length $s$ of the continued fraction of $\sqrt n$ is even, we have $\paren {-1}^{r s} = 1$.
Hence $x = p_{r s}, y = q_{r s}$ is a solution to Pell's equation for each $r \ge 1$.
When $s$ is odd, though:
:$x = p_{r s}, y = q_{r s}$ is a solution of $x^2 - n y^2 = -1$ when $r$ is odd
:$x = p_{r s}, y = q_{r s}$ is a solution of $x^2 - n y^2 = 1$ when $r$ is even.
== Sequence ==
== Examples ==
== Sources ==
* 
* 
* 
Category:Pell's Equation
Solution to Prisoner's Dilemma

== Solution to Prisoner's Dilemma ==
== Proof ==
From the payoff table:
Each player gains by cooperating, and the best outcome is for neither player to confess.
However, because each player then has the opportunity to improve his position by changing his strategy to confessing, $\tuple {\text {Don't Confess}, \text {Don't Confess} }$ is not a Nash equilibrium.
Thus there is a single Nash equilibrium:
:$\tuple {\text {Confess}, \text {Confess} }$
== Sources ==
* : $\text I$ Strategic Games: Chapter $2$ Nash Equilibrium: $2.3$: Examples: Example $16.2$
Category:Prisoner's Dilemma
Solution to Quadratic Congruence

== Theorem ==
The quadratic congruence:
:$a x^2 + b x + c \equiv 0 \pmod n$
can be solved by solving the congruence:
:$y^2 \equiv \paren {b^2 - 4 a c} \pmod n$
and the linear congruence:
:$2 a x + b \equiv y \pmod n$
== Proof ==
== Sources ==
* 
* 
Category:Quadratic Congruences
Solution to Quadratic Equation

== Theorem ==
The quadratic equation of the form $a x^2 + b x + c = 0$ has solutions:
:$x = \dfrac {-b \pm \sqrt {b^2 - 4 a c} } {2 a}$
=== Real Coefficients ===
== Proof ==
== Also known as ==
== Sources ==
* : Chapter $\text I$ Introductory: $2$. The Use of Symbols
* : $\S 1.2$. The Algebraic Theory: $(1.20)$
* : $3$: Elementary Analytic Methods: $3.8$ Algebraic Equations: Solution of Quadratic Equations: $3.8.1$
* : $\S 9$: Solutions of Algebraic Equations: Quadratic Equation: $a x^2 + b x + c = 0$: $9.1$
* : $\S 1$: Real Numbers: $\S 1.10$: Quadratic equations
* : $1$: Complex Numbers: Solved Problems: Polynomial Equations: $31$
* 
* : Where to begin...
* 
* : Chapter $4$: Lure of the Unknown: Equations
* : $\S 5$: Solutions of Algebraic Equations: Quadratic Equation: $a x^2 + b x + c = 0$: $5.1.$
* 
Category:Quadratic Equations
Solution to Quadratic Equation/Real Coefficients

== Theorem ==
Let $a, b, c \in \R$.
The quadratic equation $a x^2 + b x + c = 0$ has:
:Two real solutions if $b^2 - 4 a c > 0$
:One real solution if $b^2 - 4 a c = 0$
:Two complex solutions if $b^2 - 4 a c < 0$, and those two solutions are complex conjugates.
== Proof ==
From Solution to Quadratic Equation:
:$x = \dfrac {-b \pm \sqrt {b^2 - 4 a c} } {2 a}$
If the discriminant $b^2 - 4 a c > 0$ then $\sqrt {b^2 - 4 a c}$ has two values and the result follows.
If the discriminant $b^2 - 4 a c = 0$ then $\sqrt {b^2 - 4 a c} = 0$ and $x = \dfrac {-b} {2 a}$.
If the discriminant $b^2 - 4 a c < 0$, then it can be written as:
:$b^2 - 4 a c = \paren {-1} \size {b^2 - 4 a c}$
Thus:
:$\sqrt {b^2 - 4 a c} = \pm i \sqrt {\size {b^2 - 4 a c} }$
and the two solutions are:
:$x = \dfrac {-b} {2 a} + i \dfrac {\sqrt {\size {b^2 - 4 a c} } } {2 a}, x = \dfrac {-b} {2 a} - i \dfrac {\sqrt {\size {b^2 - 4 a c} } } {2 a}$
and once again the result follows.
== Sources ==
* : $3$: Elementary Analytic Methods: $3.8$ Algebraic Equations: Solution of Quadratic Equations: $3.8.1$
* : $\S 9$: Solutions of Algebraic Equations: Quadratic Equation: $a x^2 + b x + c = 0$: $9.1$
* : $\S 1$: Real Numbers: $\S 1.10$: Quadratic equations
* 
* 
* : $\S 5$: Solutions of Algebraic Equations: Quadratic Equation: $a x^2 + b x + c = 0$: $5.1.$
* 
Category:Quadratic Equations
Solution to Separable Differential Equation

== Definition ==
Consider the separable differential equation:
:$\dfrac {\d y} {\d x} = \map g x \map h y$
Its general solution is found by solving the integration:
:$\ds \int \frac {\d y} {\map h y} = \int \map g x \rd x + C$
This technique is generally known as Separation of Variables.
=== General Result ===
== Proof ==
Dividing both sides by $\map h y$, we get:
:$\dfrac 1 {\map h y} \dfrac {\d y} {\d x} = \map g x$
Integrating both sides $x$, we get:
:$\ds \int \frac 1 {\map h y} \frac {\d y} {\d x} \rd x = \int \map g x \rd x$
which, from Integration by Substitution, reduces to the result.
The arbitrary constant $C$ appears during the integration process.
== Mnemonic Device ==
As derivatives are not fractions, the following is a only. 
This is an an abuse of notation that is likely to make some Calculus professors upset. 
But it's useful. 
== Examples ==
== Sources ==
* : $\S 2.7$: Homogeneous Equations
* : : $(2)$ Variables separable
* : : $(2)$ Variables separable
* 
* 
Category:Solution to Separable Differential Equation
Category:Separable Differential Equations
Solution to Separable Differential Equation/Examples/Arbitrary Example 1

== Example of Use of Solution to Separable Differential Equation ==
Consider the first order ODE:
:$(1): \quad \map {\dfrac \d {\d x} } {\map f x} = 3 x$
where we are given that $\map f 1 = 2$.
The particular solution to $(1)$ is:
:$\map f x = \dfrac {3 x^2 + 1} 2$
== Proof ==
We use Solution to Separable Differential Equation:
When $x = 1$, we have that $\map f x = 2$.
Hence:
Hence:
:$\map f x = \dfrac {3 x^2 + 1} 2$
== Sources ==
* : $6$. Integral Calculus: Exercises: $6.1$ The Primitive Function: $1. \ \text {(a)}$
Category:Examples of Solutions to Separable Differential Equation
Solution to Separable Differential Equation/General Result

== Theorem ==
Consider the separable differential equation:
:$\map {g_1} x \map {h_1} y + \map {g_2} x \map {h_2} y \dfrac {\d y} {\d x} = 0$
Its general solution is found by solving the integration:
:$\ds \int \frac {\map {g_1} x} {\map {g_2} x} \rd x + \int \frac {\map {h_2} y} {\map {h_1} y} \rd y = C$
== Proof ==
== Sources ==
* : $\S 18$: Basic Differential Equations and Solutions: $18.1$: Separation of variables
* : $\S 19$: Basic Differential Equations and Solutions: $19.1.$: Separation of variables
Category:Solution to Separable Differential Equation