nyuuzyou/wikis huggingface dataset:
Basis of Point Lattice is not Necessarily Unique

== Theorem ==
Let $\LL$ be a point lattice of dimension $n$.
Let $\BB = \set {\mathbf v_1, \mathbf v_2, \ldots, \mathbf v_n}$ be a basis of $\LL$.
Then it is not necessarily the case that $\BB$ is the only basis of $\LL$.
== Proof ==
== Sources ==
* 
* 
Category:Bases of Point Lattices
Category:Point Lattices
Basis of Vector Space Injects into Generator

== Theorem ==
Let $K$ be a division ring.
Let $V$ be a vector space over $K$.
Let $B$ be a basis of $V$.
Let $G$ be a generator of $V$.
Then there exists an injection from $B$ to $G$.
== Proof ==
By Vector Space has Basis between Linearly Independent Set and Spanning Set, there exists a basis $C \subset G$.
By Bases of Vector Space have Equal Cardinality, there exists a bijection between $B$ and $C$.
By Composite of Injections is Injection, composing this bijection with the inclusion of $C$ in $G$, we obtain an injection from $B$ to $G$.
Category:Vector Spaces
Basis of Vector Space of Polynomial Functions

== Theorem ==
Let $B$ be the set of all the identity functions $I^n$ on $\R^n$ where $n \in \N^*$.
Then $B$ is a basis of the $\R$-vector space $\map P \R$ of all polynomial functions on $\R$.
== Proof ==
By definition, every real polynomial function is a linear combination of $B$.
Suppose:
:$\ds \sum_{k \mathop = 0}^m \alpha_k I^k = 0, \alpha_m \ne 0$
Then by differentiating $m$ times, we obtain from Nth Derivative of Nth Power:
:$m! \alpha_m = 0$
whence $\alpha_m = 0$ which is a contradiction.
Hence $B$ is linearly independent and therefore is a basis for $\map P \R$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Example $27.7$
Category:Bases of Vector Spaces
Category:Polynomial Theory
Bayes' Theorem

== Theorem ==
Let $\Pr$ be a probability measure on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $\condprob A B$ denote the conditional probability of $A$ given $B$.
Let $\map \Pr A > 0$ and $\map \Pr B > 0$.
Then:
:$\condprob B A = \dfrac {\condprob A B \, \map \Pr B} {\map \Pr A}$
=== General Result ===
There are other more or less complicated ways of saying very much the same thing, all of which can be derived from the basic version with the help of other fairly elementary results.
For example:
== Proof ==
From the definition of conditional probabilities, we have:
:$\condprob A B = \dfrac {\map \Pr {A \cap B} } {\map \Pr B}$
:$\condprob B A = \dfrac {\map \Pr {A \cap B} } {\map \Pr A}$
After some algebra:
:$\condprob A B \, \map \Pr B = \map \Pr {A \cap B} = \condprob B A \, \map \Pr A$
Dividing both sides by $\map \Pr A$ (we are told that it is non-zero), the result follows:
:$\condprob B A = \dfrac {\condprob A B \, \map \Pr B} {\map \Pr A}$
== Also known as ==
== Examples ==
== Historical Note ==
== Sources ==
* : $\S 1.6$: Exercise $18$
* : $1.2$ Basic Concepts of Decision Theory
Category:Bayes' Theorem
Category:Bayesian Decision Theory
Category:Probability Theory
Bayes' Theorem/Also known as

== Bayes' Theorem: Also known as ==
This result is also known as Bayes' Formula.
The formula:
:$\condprob A B \, \map \Pr B = \map \Pr {A \cap B} = \condprob B A \, \map \Pr A$
is sometimes called the product rule for probabilities.
 
== Sources ==
* : $1.2$ Basic Concepts of Decision Theory
Category:Bayes' Theorem
Bayes' Theorem/Examples/Arbitrary Example 1

== Example of Use of Bayes' Theorem ==
Let box $1$ contain $10$ good screws and $2$ unslotted screws.
Let box $2$ contain $8$ good screws and $4$ unslotted screws.
Let a box be selected at random.
Let a screw chosen from that box prove to be unslotted.
What is the probability that it came from box $2$?
== Solution ==
Let $A$ be the event: An unslotted screw is selected.
Let $B_1$ be the event: Screw is selected from box $1$.
Let $B_2$ be the event: Screw is selected from box $2$.
We have that:
Hence the probability is $\dfrac 2 3$ that the unslotted screw came from box $2$.
== Sources ==
* 
* 
Category:Bayes' Theorem
Bayes' Theorem/Examples/Arbitrary Example 2

== Example of Use of Bayes' Theorem ==
Suppose that, in a population, $6$ out of every $1000$ people has an illness $X$.
It is known that:
:if a person has $X$, there is a $92 \%$ probability that a blood test will be positive for $X$
:if a person does not have $X$, there is a $0 \cdotp 5 \%$ probability that a blood test will be positive for $X$.
Let a person selected at random test positive for $X$.
What is the probability that this person actually has $X$?
== Solution ==
Let $A$ be the event: Person tests positive for $X$.
Let $B_1$ be the event: Person has $X$.
Let $B_2$ be the event: Person does not have $X$.
We have that:
Hence the probability that a person has $X$ is $0 \cdotp 6 \%$, but after testing positive the probability is $52 \cdotp 6 \%$.
== Sources ==
* 
Category:Bayes' Theorem
Bayes' Theorem/General Result

== Theorem ==
Let $\Pr$ be a probability measure on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $\set {B_1, B_2, \ldots}$ be a partition of the event space $\Sigma$.
Then, for any $B_i$ in the partition:
:$\condprob {B_i} A = \dfrac {\condprob A {B_i} \map \Pr {B_i} } {\map \Pr A} = \dfrac {\condprob A {B_i} \map \Pr {B_i} } {\sum_j \condprob A {B_j} \map \Pr {B_j} }$
where $\ds \sum_j$ denotes the sum over $j$.
== Proof ==
Follows directly from the Total Probability Theorem:
:$\ds \map \Pr A = \sum_i \condprob A {B_i} \map \Pr {B_i}$
and Bayes' Theorem:
:$\condprob {B_i} A = \dfrac {\condprob A {B_i} \map \Pr {B_i} } {\map \Pr A}$
== Examples ==
== Historical Note ==
== Sources ==
* 
* 
* 
Category:Bayes' Theorem
Beatty's Theorem

== Theorem ==
Let $r, s \in \R \setminus \Q$ be an irrational number such that $r > 1$ and $s > 1$.
Let $\BB_r$ and $\BB_s$ be the Beatty sequences on $r$ and $s$ respectively.
Then $\BB_r$ and $\BB_s$ are complementary Beatty sequences :
:$\dfrac 1 r + \dfrac 1 s = 1$
== Proof 1 ==
== Proof 2 ==
== Also known as ==
Beatty's theorem is also known as the Rayleigh theorem, for .
Category:Beatty Sequences
Beatty's Theorem/Proof 1

== Theorem ==
== Proof ==
We have been given that $r > 1$.
Let $\dfrac 1 r + \dfrac 1 s = 1$.
Then:
:$s = \dfrac r {r - 1}$
It is to be shown that every positive integer lies in exactly one of the two Beatty sequences $\BB_r$ and $\BB_s$.
Consider the ordinal positions occupied by all the fractions $\dfrac j r$ and $\dfrac k s$ when they are jointly listed in nondecreasing order for positive integers $j$ and $k$.
 that $\dfrac j r = \dfrac k s$ for some $j, k \in \Z_{>0}$.
Then:
:$\dfrac r s = \dfrac j k$
which is rational.
But also:
:$\dfrac r s = r \paren {1 - \dfrac 1 r} = r - 1$
which is not rational.
Therefore, no two of the numbers occupy the same position.
Consider some $\dfrac j r$.
There are $j$ numbers $\dfrac i r \le \dfrac j r$.
There are also $\floor {\dfrac {j s} r}$ numbers $\dfrac k s \le \dfrac j r$.
So the position of $\dfrac j r$ in the list is $j + \floor {\dfrac {j s} r}$.
The equation $\dfrac 1 r + \dfrac 1 s = 1$ implies:
:$j + \floor {\dfrac {j s} r} = j + \floor {j \paren {s - 1} } = \floor {j s}$
Likewise, the position of $\dfrac k s$ in the list is $\floor {k r}$.
It is concluded that every positive integer corresponding to every position in the list is of the form $\floor {n r}$ or of the form $\floor {n r}$, but not both.
The converse statement is also true: if $p$ and $q$ are two real numbers such that every positive integer occurs precisely once in the above list, then $p$ and $q$ are irrational and the sum of their reciprocals is $1$.
Category:Beatty Sequences
Beatty's Theorem/Proof 2

== Theorem ==
== Proof ==
=== Collisions ===
 there exist integers $j > 0, k, m$ such that:
:$j = \floor {k \cdot r} = \floor {m \cdot s}$
This is equivalent to the inequalities:
:$j \le k \cdot r < j + 1$
and:
:$j \le m \cdot s < j + 1$
As $r$ and $s$ are irrational, equality cannot happen.
So:
:$j < k \cdot r < j + 1$
and:
:$j < m \cdot s < j + 1$
which leads to:
:$\dfrac j r < k < \dfrac {j + 1} r$
and:
:$\dfrac j s < m < \dfrac {j + 1} s$
Adding these together and using the by hypothesis:
:$j < k + m < j + 1$
Thus there is an integer strictly between two adjacent integers.
This is impossible.
Thus the supposition must be false.
=== Anti-collisions ===
 that there exist integers $j > 0, k, m$ such that:
:$k \cdot r < j$
and:
:$j + 1 \le \paren {k + 1} \cdot r$
and:
:$m \cdot s < j$
and:
:$j + 1 \le \paren {m + 1} \cdot s$
Since $j + 1 \ne 0$, and $r$ and $s$ are irrational, equality cannot happen.
So:
:$k \cdot r < j$
and:
:$j + 1 < \paren {k + 1} \cdot r$
and:
:$m \cdot s < j$
and:
:$j + 1 < \paren {m + 1} \cdot s$
Then:
:$k < \dfrac j r$:
and:
:$\dfrac {j + 1} r < k + 1$
and:
:$m < \dfrac j s$
and:
:$\dfrac {j + 1} s < m + 1$
Adding corresponding inequalities:
:$k + m < j$
and:
:$j + 1 < k + m + 2$
:$k + m < j < k + m + 1$
which is also impossible.
Thus the supposition is false.
Category:Beatty Sequences
Beer-Lambert-Bouguer Law

== Physical Law ==
The fraction of incident light absorbed by a thin layer of translucent material is proportional to the thickness of the layer.
== Also known as ==
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 4$: Growth, Decay and Chemical Reactions: Problem $6$
Category:Beer-Lambert-Bouguer Law
Category:Optics
Beer-Lambert-Bouguer Law/Also known as

== Beer-Lambert-Bouguer Law: Also known as == 
The Beer-Lambert-Bouguer Law is also known as:
:the Lambert-Beer Law
:Beer's Law
:the Beer-Lambert Law
:Lambert's Law of Absorption
:Lambert's Law.
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 4$: Growth, Decay and Chemical Reactions: Problem $6$
Category:Beer-Lambert-Bouguer Law
Begging the Question

== Fallacy ==
Let $p$ be a proposition whose truth value is to be proved (either true or false).
Suppose a logical argument is created which uses $p$ as a premise.
That is, that the proposition to be derived is used to derive itself.
Then the fallacy of begging the question has been perpetrated.
== Also see ==
* Denying the Antecedent
* Affirming the Consequent
== Linguistic Note ==
== Sources ==
* : $\text{I}$: 'NOT' and 'IF': $\S 4$
Category:Fallacies
Begging the Question/Linguistic Note

== Linguistic Note on Begging the Question ==
The term Begging the Question originated as a 16th century mistranslation of Latin petitio principii, which means assuming the initial point.
Consequently there is considerable confusion in natural language as to exactly what this phrase actually means.
In contemporary and colloquial language it is commonly used to mean raising the question, in particular to call attention to a question or a point for debate which has conspicuously not been raised but, in the questioner's mind, ought to have been, as has been made apparent during the process of the argument under discussion.
The cliché "elephant in the room" can frequently be seen in close proximity, and should arguably be used instead.
Category:Linguistic Notes
Behaviour of Function Near Limit

== Theorem ==
Let $f$ be a real function.
Let $f \left({x}\right) \to l$ as $x \to \xi$.
Then:
:If $l > 0$, then $\exists h > 0: \forall x: \xi - h < x < \xi + h, x \ne \xi: \map f x > 0$
:If $l < 0$, then $\exists h > 0: \forall x: \xi - h < x < \xi + h, x \ne \xi: \map f x < 0$
== Proof ==
From the definition of limit of a function:
:$\forall \epsilon > 0: \exists \delta > 0: 0 < \size {x - \xi} < \delta \implies \size {\map f x - l} < \epsilon$
Let $l > 0$.
Since this is true for all $\epsilon > 0$, it is also true for $\epsilon = l$.
So let the value of $\delta$, for the above to be true, labelled $h$.
Then:
:$0 < \size {x - \xi} < h \implies \size {\map f x - l} < l$
That is:
:$\xi - h < x < \xi + h, x \ne \xi \implies 0 = l - l < \map f x < l + l = 2 l$
Hence:
:$\forall x: \xi - h < x < \xi + h, x \ne \xi: 0 < \map f x$
Now let $l < 0$, and consider $\epsilon = -l$.
A similar thread of reasoning leads us to:
:$\xi - h < x < \xi + h, x \ne \xi \implies -2 l < \map f x < 0$
and hence the second result.
== Note ==
If $l = 0$, neither conclusion may be drawn without further information - $\map f x$ may stay either side of zero.
== Sources ==
* : $\S 8.15 \ (6)$
Category:Limits of Real Functions
Behaviour of Parametric Equations for Folium of Descartes according to Parameter

== Theorem ==
Consider the folium of Descartes $F$, given in parametric form as:
:$\begin {cases} x = \dfrac {3 a t} {1 + t^3} \\ y = \dfrac {3 a t^2} {1 + t^3} \end {cases}$
Then:
:$F$ has a discontinuity at $t = -1$.
:For $t < -1$, the section in the $4$th quadrant is generated
:For $-1 < t \le 0$, the section in the $2$nd quadrant is generated 
:For $0 \le t$, the section in the $1$st quadrant is generated.
== Proof ==
:
=== Discontinuity at $t = -1$ ===
When $t = -1$, we have that:
:$1 + t^3 = 0$
and so both $x$ and $y$ are undefined.
=== Behaviour for $t < -1$ ===
So:
:$\ds \lim_{t \mathop \to -\infty} \tuple {x, y} = \tuple {0^+, 0^-}$
Then:
Thus:
Thus, as $t$ goes from $-\infty$ to $-1$, $F$ goes from $\tuple {0, 0}$ in the $4$th quadrant to $\tuple {+\infty, -\infty}$.
=== Behaviour for $t > 0$ ===
So:
:$\ds \lim_{t \mathop \to -\infty} \tuple {x, y} = \tuple {0^+, 0^+}$
It is observed that for $t > 0$, we have that:
:$x > 0$
:$y > 0$
and so for $t > 0$, $F$ is in the $1$st quadrant.
Thus we have that the loop is traversed from $t = 0$ to $t = +\infty$.
=== Behaviour for $-1 < t < 0$ ===
We have that when $t = 0$, $\tuple {x, y} = \tuple {0, 0}$.
When $-1 < t < 0$, we have that:
Then:
Thus:
Thus, as $t$ goes from $0$ to $-1$, $F$ goes from $\tuple {0, 0}$ in the $2$nd quadrant to $\tuple {-\infty, +\infty}$.
The result follows.
Category:Folium of Descartes
Bell Number/Examples/B3

== Example of Bell Number ==
From the definition of Bell number, $B_3$ is the number of ways a set with $3$ elements can be partitioned.
:$\set {a, b, c}$
:$\set {\set a, \set {b, c} }$
:$\set {\set b, \set {a, c} }$
:$\set {\set c, \set {a, b} }$
:$\set {\set a, \set b, \set c}$
:$B_3 = 5$
Category:Examples of Bell Numbers
Bell Number as Summation over Lower Index of Stirling Numbers of the Second Kind

== Theorem ==
Let $B_n$ be the Bell number for $n \in \Z_{\ge 0}$.
Then:
:$B_n = \ds \sum_{k \mathop = 0}^n {n \brace k}$
where $\ds {n \brace k}$ denotes a Stirling number of the second kind.
=== Corollary ===
When $n > 0$, the slightly simpler form can be used:
== Proof ==
By definition of Bell numbers:
:$B_n$ is the number of partitions of a (finite) set whose cardinality is $n$.
First consider the case where $n > 0$.
From Number of Set Partitions by Number of Components, the number of partitions of $S$ into $k$ components is $\ds {n \brace k}$.
Thus the total number of all partitions of $S$ is the sum of $\ds {n \brace k}$ where $k$ runs from $1$ to $n$.
Hence:
:$B_n = \ds \sum_{k \mathop = 1}^n {n \brace k}$
But $\ds {n \brace 0} = 0$ for all $n \in \Z_{> 0}$, and so:
:$B_n = \ds \sum_{k \mathop = 0}^n {n \brace k}$
Now consider the edge case where $n = 0$, that is, $S = \O = \set{}$.
There is by conventional definition one partition of $\O$, that is: $\set \O$
In this case, we have:
:$\ds {0 \brace 0} = 1$
and so:
:$B_0 = \ds \sum_{k \mathop = 0}^0 {n \brace k} = {0 \brace 0} = 1$
Category:Bell Numbers
Category:Stirling Numbers
Bell Number as Summation over Lower Index of Stirling Numbers of the Second Kind/Corollary

== Theorem ==
Let $B_n$ be the Bell number for $n \in \Z_{> 0}$.
Then:
:$B_n = \ds \sum_{k \mathop = 1}^n {n \brace k}$
where $\ds {n \brace k}$ denotes a Stirling number of the second kind.
== Proof ==
From Bell Number as Summation over Lower Index of Stirling Numbers of the Second Kind, we have that:
:$B_n = \ds \sum_{k \mathop = 0}^n {n \brace k}$
But when $n > 0$:
:$\ds {n \brace 0} = 0$
Hence the result.
Category:Bell Numbers
Category:Stirling Numbers
Belphegor's Prime

== Interesting Number ==
Belphegor's prime is the palindromic prime:
:$1000000000000066600000000000001$
It has:
:$666$: the Number of the Beast in its center
:Unlucky number $13$ zeroes either side of the $666$
:$31$ digits, which is not only itself prime, but backwards is $13$ which is also prime.
== Linguistic Note ==
Category:Specific Numbers
Category:Palindromic Primes
Bendixson-Dulac Theorem

== Theorem ==
Suppose there exists a continuously differentiable function $\alpha \left({x, y}\right)$ on a simply connected domain.
Suppose that:
:$\nabla \cdot \left({\alpha F}\right)$ 
is either always positive or always negative.
Then the two-dimensional autonomous system:
:$ \left({x, y}\right)' = F \left({x, y}\right)$
does not have a periodic solution.
== Proof ==
Category:Differential Equations
Benise's Theorem

== Theorem ==
</onlyinclude>
Let $x$ be a polynomial variable with exponent $n$.
Then:
:$\dfrac \d {\d x} \size {x^n} = n \cdot \dfrac {\size {x^n} } x$
for $x \ne 0$.
At $x = 0$, $\size x$ is not differentiable.
</onlyinclude>
== Proof == 
For \( x \ne 0 \):
\[
x^{n-1} = \frac{x^n}{x}
\]
For any real \( x \):
\[
x = \operatorname{sign}(x) \cdot \size{x}
\]
Beppo Levi's Theorem

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\sequence {f_n}_{n \mathop \in \N} \in \MM_{\overline \R}^+$ be an increasing sequence of positive $\Sigma$-measurable functions.
Let $\ds \sup_{n \mathop \in \N} f_n: X \to \overline \R$ be the pointwise supremum of $\sequence {f_n}_{n \mathop \in \N}$, where $\overline \R$ denotes the extended real numbers.
Then:
:$\ds \int \sup_{n \mathop \in \N} f_n \rd \mu = \sup_{n \mathop \in \N} \int f_n \rd \mu$
where the supremum on the is in the ordering on $\overline \R$.
== Proof ==
Since by definition $\ds \sup _{n \mathop \in \N} f_n \ge f_m$ for all $m$, we have:
:$\ds \int \sup_{n \mathop \in \N} f_n \rd \mu \ge \int f_m \rd \mu$
and hence the inequality holds for the supremum as well:
:$\ds \int \sup_{n \mathop \in \N} f_n \rd \mu \ge \sup_{m \mathop \in \N} \int f_m \rd \mu$
It remains to show the reverse inequality.
We have that the integral of $\sup \limits_{n \mathop \in \N}f_n$ is defined as the supremum of the integrals of positive simple functions:
:$s \le \sup \limits_{n \mathop \in \N} f_n$
Let $s$ be a simple function:
:$\ds s = \sum_{i \mathop = 1}^k \lambda_i \chi_{E_i} \le \sup_{n \mathop \in \N} f_n$
where $\lambda_i \in \closedint 0 {+\infty}$ and $E_i \in \Sigma$.
We are to show that:
:$\ds \sup_{m \mathop \in \N} \int f_m \rd \mu \ge \int s \rd \mu$
To show this, we use the fact that $\nu_s: \Sigma \to \closedint 0 {+\infty}$ defined by $\map {\nu_s} E = \ds \int \chi_Es \rd \mu$ clearly defines a measure over $X$, because it is simply a linear combination (with positive coefficients) of the measures $\bigvalueat \mu {E_i}$.
Now, if we fix $1 > \epsilon > 0$, we have that the sets:
:$A_m = \set {x \in X: \map {f_m} x \ge \paren {1 - \epsilon} \map s x}$
form a cover of $X$ ($X = \ds \bigcup_{m \mathop \in \N} A_m$) by definition of the supremum, because $\ds s \le \sup_{n \mathop \in \N} f_n$.
Furthermore, the increasing sequence $\sequence {A_m}$ has the limit $A_m \uparrow X$ ($m \to \infty$).
By definition of the $A_m$ we have that:
:$\ds \int f_m \rd \mu \ge \int \chi_{A_m} f_m \rd \mu \ge \paren {1 - \epsilon} \int \chi_{A_m} s \rd \mu = \paren {1 - \epsilon} \map {\nu_s} {A_m}$
where the first inequality follows from the fact that the $f_m$ are positive.
We have that the sequence $\sequence {f_n}_{n \mathop \in \N}$ increases monotonically.
By Measure of Limit of Increasing Sequence of Measurable Sets given in any measure space, we have that taking the supremum of both sides yields:
Since $\epsilon$ was selected arbitrarily, we have that the desired inequality holds:
:$\ds \sup_{m \mathop \in \N} \int f_m \rd \mu \ge \int s \rd \mu$
== Also known as ==
Some authors refer to this result as Beppo Levi's lemma, while others call it the monotone convergence theorem.
On the latter name is reserved for the general result: Monotone Convergence Theorem (Measure Theory).
== Sources ==
* 
* : $9.6$
Category:Measure Theory
Berlin Papyrus 6619/Examples/Problem 1

== Problem 1 from ==
:If it is said to thee:
::the area of a square of $100$ is equal to that of two smaller squares.
::The side of one is $\frac 1 2 + \frac 1 4$ the side of the other.
:Let me know the sides of the two unknown squares.
== Solution ==
The sides of the two unknown squares are $6$ and $8$.
== Modern Proof ==
== Historic Proof ==
== Sources ==
* : $14$: Equations of the First and Second Degree: Equations of the Second Degree
* : Squares Without Pythagoras: $11$
Category:Berlin Papyrus
Berlin Papyrus 6619/Examples/Problem 1/Historic Proof

== Problem 1 from ==
== Proof ==
The method of solution used is the rule of false position.
Assume that:
But the result should be:
:$100 = 64 \times \paren {1 + \dfrac 9 {16} }$
Therefore, our two squares must be $64$ times bigger.
Therefore their sides must be $8$ times bigger.
So the result is:
Category:Berlin Papyrus
Berlin Papyrus 6619/Examples/Problem 1/Modern Proof

== Problem 1 from ==
== Proof ==
Let $x$ be the length of the side of the smaller square of the two.
We have:
As this is the side of an actual square we are investigating here, we can dispose of the negative square root.
Thus we have:
:$y = 8$
and so:
:$x = \dfrac 3 4 \times 8 = 6$
Category:Berlin Papyrus
Bernoulli's Equation/2 x y dx + (x^2 + 2 y) dy = 0

== Theorem ==
The first order ODE:
:$(1): \quad 2 x y \rd x + \paren {x^2 + 2 y} \rd y = 0$
has the solution:
:$y \paren {x^2 + y} = C$
== Proof ==
It can be seen that $(2)$ is in the form:
:$\dfrac {\d x} {\d y} + \map P y x = \map Q y x^n$
where:
:$\map P y = \dfrac 1 {2 y}$
:$\map Q y = -1$
:$n = -1$
and so is an example of Bernoulli's equation.
By Solution to Bernoulli's Equation it has the general solution:
:$(3): \quad \ds \frac {\map \mu y} {x^{n - 1} } = \paren {1 - n} \int \map Q y \, \map \mu y \rd y + C$
where:
:$\map \mu y = e^{\paren {1 - n} \int \map P y \rd y}$
Thus $\map \mu x$ is evaluated:
and so substituting into $(3)$:
Category:Examples of Bernoulli's Equation
Bernoulli's Equation/x^2 dy = (2 x y + y^2) dx

== Theorem ==
The first order ODE:
:$(1): \quad x^2 \rd y = \paren {2 x y + y^2} \rd x$
has the general solution:
:$y = -\dfrac {x^2} {x + C}$
== Proof ==
Rearranging $(1)$:
:$(2): \quad \dfrac {\d y} {\d x} - \dfrac 2 x y = \dfrac {y^2} {x^2}$
It can be seen that $(2)$ is in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$
where:
:$\map P x = -\dfrac 2 x$
:$\map Q x = \dfrac 1 {x^2}$
:$n = 2$
and so is an example of Bernoulli's equation.
By Solution to Bernoulli's Equation it has the general solution:
:$(3): \quad \ds \frac {\map \mu x} {y^{n - 1} } = \paren {1 - n} \int \map Q x \map \mu x \rd x + C$
where:
:$\map \mu x = e^{\paren {1 - n} \int \map P x \rd x}$
Thus $\map \mu x$ is evaluated:
and so substituting into $(3)$:
Category:Examples of Bernoulli's Equation
Bernoulli's Equation/x y' + y = x^4 y^3

== Theorem ==
The first order ODE:
:$(1): \quad x y' + y = x^4 y^3$
has the general solution:
:$\dfrac 1 {y^2} = - x^4 + C x^2$
== Proof ==
Let $(1)$ be rearranged as:
:$(2): \quad \dfrac {\d y} {\d x} + \dfrac 1 x y = x^3 y^3$
It can be seen that $(2)$ is in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$
where:
:$\map P x = \dfrac 1 x$
:$\map Q x = x^3$
:$n = 3$
and so is an example of Bernoulli's equation.
By Solution to Bernoulli's Equation it has the general solution:
:$(3): \quad \ds \frac {\map \mu x} {y^{n - 1} } = \paren {1 - n} \int \map Q x \map \mu x \rd x + C$
where:
:$\map \mu x = e^{\paren {1 - n} \int \map P x \rd x}$
Thus $\map \mu x$ is evaluated:
and so substituting into $(3)$:
Hence the general solution to $(1)$ is:
:$\dfrac 1 {y^2} = - x^4 + C x^2$
== Sources ==
* : $\S 2.10$: Problem $3 \ \text{(a)}$
Category:Examples of Bernoulli's Equation