nyuuzyou/wikis huggingface dataset:
Different Representations to Number Base represent Different Integers

== Theorem ==
Let $k \in \Z$ such that $k \ge 2$.
Let $a$ and $b$ be representations of integers in base $k$ notation:
:$a = \ds \sum_{j \mathop = 0}^r a_j k^j$
:$b = \ds \sum_{j \mathop = 0}^s b_j k^j$
such that either:
:$r \ne s$
or:
:$\exists j \in \set {0, 1, \ldots, r}: a_j \ne b_j$
Then $a$ and $b$ represent different integers.
== Proof ==
First suppose that $r \ne s$.
, suppose $r > s$.
Then from Bounds for Integer Expressed in Base k:
Otherwise $r = s$.
Let $l$ be the largest such that $a_l \ne b_l$.
, suppose $a_l > b_l$.
For all $j > l$, let $a_1 = a - a_j k_j$ and $b_1 = b - b_j k_j$.
As $a_j = b_j$ in this range, $a_j k_j = b_j k_j$ and so the same amount is being subtracted from both.
So consider $\paren {a_l - b_l} k_l$.
From Bounds for Integer Expressed in Base k:
:$\paren {a_l - b_l} k_l > \ds \sum_{j \mathop = 0}^{l - 1} a_j k^j$
and:
:$\paren {a_l - b_l} k_l > \ds \sum_{j \mathop = 0}^{l - 1} b_j k^j$
and so $a_1 > b_1$.
Hence:
:$a_1 + a_j k_j > b_1 + b_j k_j$
Hence:
:$a \ne b$
and hence the result.
== Sources ==
* : $\text {1-2}$ The Basis Representation Theorem: Exercise $7$
Category:Number Bases
Differentiability Class/Examples/Class 0 Function with Derivative Discontinuous at Point

== Example of Differentiability Class ==
Let $f$ be the real function defined as:
:$\map f x = \begin {cases} x^2 \sin \dfrac 1 x & : x \ne 0 \\ 0 & : x = 0 \end {cases}$
Then $f \in C^0$ but $f \notin C^1$.
== Proof ==
For $x = 0$:
For $x \ne 0$:
From Differentiable Function is Continuous, $f \in C^0$.
$f'$ is not continuous at $0$, so $f \notin C^1$.
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.2$ Derivatives
Category:Examples of Differentiability Classes
Differentiability Class/Examples/Class 1 Function

== Example of Differentiability Class ==
Let $f$ be the real function defined as:
:$\map f x = \begin {cases} 0 & : x < 0 \\ x^2 & : x \ge 0 \end {cases}$
Then $f \in C^1$ but $f \notin C^2$. 
== Proof ==
By inspection it is seen that $f$ is continuous everywhere.
We have that:
:$\map {f'} x = \begin {cases} 0 & : x < 0 \\ 2 x & : x \ge 0 \end {cases}$
By inspection it is seen that $f'$ is continuous everywhere.
So by definition of differentiability class, $f$ is a member of $C^1$.
Then we have that:
:$\map {f} x = \begin {cases} 0 & : x < 0 \\ 2 & : x \ge 0 \end {cases}$
Hence $f$ is not continuous at $x = 0$.
So by definition of differentiability class, $f$ is not a member of $C^2$.
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.2$ Derivatives: Example $\text F$
Category:Examples of Differentiability Classes
Differentiability Class/Examples/Class Zero Function

== Example of Differentiability Class ==
Let $f$ be the real function defined as:
:$\map f x = \begin {cases} 0 & : x < 0 \\ x & : x \ge 0 \end {cases}$
Then $f \in C^0$ but $f \notin C^1$. 
== Proof ==
By inspection it is seen that $f$ is continuous everywhere.
We have that:
:$\map {f'} x = \begin {cases} 0 & : x < 0 \\ 1 & : x \ge 0 \end {cases}$
Hence $f'$ is not continuous at $x = 0$.
So by definition of differentiability class, $f$ is not a member of $C^1$.
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.2$ Derivatives: Example $\text F$
Category:Examples of Differentiability Classes
Differentiability Class/Examples/Class n Function

== Example of Differentiability Class ==
Let a real function $f$ be required that has the following properties:
:$(1): \quad f \in C^n$
:$(2): \quad f \notin C^{n + 1}$
where $C^k$ denotes the differentiability class of order $k$.
Then $f$ may be defined as:
:$\map f x = \begin {cases} 0 & : x < 0 \\ x^{n + 1} & : x \ge 0 \end {cases}$
== Proof ==
By inspection it is seen that $f$ is continuous everywhere.
We have from Nth Derivative of Mth Power that:
:$\map {f^{\paren n} } x = \begin {cases} 0 & : x < 0 \\ \paren {n + 1}^{\underline n} x & : x \ge 0 \end {cases}$
where $\paren {n + 1}^{\underline n}$ denotes the $n$th falling factorial of $n + 1$.
At $x = 0$ it is seen that $\map {f^{\paren n} } x = 0$ and so $\map {f^{\paren n} } x$ is continuous at $0$.
By inspection it is seen that $f^{\paren n}$ is continuous everywhere.
So by definition of differentiability class, $f$ is a member of $C^n$.
Then we have that:
:$\map {f^{\paren {n + 1} } } x = \begin {cases} 0 & : x < 0 \\ \paren {n + 1}! & : x \ge 0 \end {cases}$
Hence $f^{\paren {n + 1} }$ is not continuous at $x = 0$.
So by definition of differentiability class, $f$ is not a member of $C^{n + 1}$.
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.2$ Derivatives
Category:Examples of Differentiability Classes
Differentiability Class is Subset of Differentiability Class of Lower Order

== Theorem ==
Let $f$ be a real function. 
Let $f$ be an element of differentiability class $C^n$.
Then:
:$\forall k \in \set {0, 1, \ldots, n - 1}: f \in C^k$
That is, $f$ is in all differentiability classes of order less than $n$.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
:$f \in C^n \implies \forall k \in \set {0, 1, \ldots, n - 1}: f \in C^k$
=== Basis for the Induction ===
$\map P 1$ is the case:
:$f \in C^1 \implies f \in C^0$
By definition of differentiability class $C^0$:
:$C^0$ is the class of continuous real functions $C$.
Hence $\map P 1$ is seen to hold from Function in Differentiability Class 1 is also in Continuity Class:
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$f \in C^r \implies \forall k \in \set {0, 1, \ldots, r - 1}: f \in C^k$
from which it is to be shown that:
:$f \in C^{r + 1} \implies \forall k \in \set {0, 1, \ldots, r }: f \in C^k$
=== Induction Step ===
This is the induction step:
Let $f \in C^{r + 1}$.
Let $g = \dfrac {\d^r} {\d x^r} \map f x$.
Then:
:$\dfrac {\d g} {\d x} = \dfrac {\d^{r + 1} } {\d x^{r + 1} } \map f x$
Thus $g \in C^1$.
By the basis for the induction it follows that $g \in C^0$.
That is, $g$ is continuous.
Hence by definition of $g$, $\dfrac {\d^r} {\d x^r} \map f x$ is continuous.
That is:
:$f \in C^r$
By the induction hypothesis:
:$\forall k \in \set {0, 1, \ldots, r - 1}: f \in C^k$
It follows that:
:$\forall k \in \set {0, 1, \ldots, r}: f \in C^k$
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: f \in C^n \implies \forall k \in \set {0, 1, \ldots, n - 1}: f \in C^k$
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.2$ Derivatives
Category:Differentiability Classes
Differentiability of Function with Translation Property

== Theorem ==
Let $f$ be a real function.
Let $f$ have the translation property.
Let $c$ be a real number.
Let $\map {f'} c$ exist.
Then:
:$f'$ exists
:$f'$ is a constant function
== Proof ==
Let $x$ be a real number.
We have:
We conclude that $\map {f'} x$ exists as $\map {f'} c$ exists.
Also, $\map {f'} x$ exists everywhere as $x$ is arbitrary.
In other words, $f'$ exists.
Also, $f'$ is a constant function as $\map {f'} x$ equals $\map {f'} c$ for every real number $x$.
Category:Differential Calculus
Differentiability under Integral Sign

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\openint a b$ be a non-empty open interval.
Let $f: \openint a b \times X \to \R$ be a mapping satisfying:
:$(1): \quad$ For all $t \in \openint a b$, the mapping $x \mapsto \map f {t, x}$ is $\mu$-integrable
:$(2): \quad$ For all $x \in X$, the mapping $t \mapsto \map f {t, x}$ is differentiable
:$(3): \quad$ There exists a $\mu$-integrable $g: X \to \R$ such that:
::::$\ds \forall \tuple {t, x} \in \openint a b \times X: \size {\frac \partial {\partial t} \map f {t, x} } \le \map g x$
Then the mapping $h: \openint a b \to \R$ defined by:
:$\ds \map h t := \int \map f {t, x} \map {\rd \mu} x$
is differentiable, and its derivative is:
:$\ds \dfrac \d {\d t} \map h t = \int \frac \partial {\partial t} \map f {t, x} \map {\rd \mu} x$
== Proof ==
== Sources ==
* : $11.5$
Category:Measure Theory
Differentiable Bounded Concave Real Function is Constant

== Theorem ==
Let $f$ be a real function which is:
:$(1): \quad$ differentiable on $\R$
:$(2): \quad$ bounded on $\R$
:$(3): \quad$ concave on $\R$.
Then $f$ is constant.
== Proof ==
Let $f$ be differentiable and bounded on $\R$.
Let $f$ be concave on $\R$.
Let $\xi \in \R$.
 $\map {f'} \xi > 0$.
Then by Mean Value of Concave Real Function it follows that:
:$\map f x \le \map f \xi + \map {f'} \xi \paren {x - \xi} \to -\infty$ as $x \to -\infty$
and therefore is not bounded.
Similarly, suppose $\map {f'} \xi < 0$.
Then by Mean Value of Concave Real Function it follows that:
:$\map f x \le \map f \xi + \map {f'} \xi \paren {x - \xi} \to -\infty$ as $x \to +\infty$
and therefore is likewise not bounded.
Hence $\map {f'} \xi = 0$.
From Zero Derivative implies Constant Function, it follows that $f$ is constant.
== Also see ==
* Differentiable Bounded Convex Real Function is Constant
Category:Concave Real Functions
Category:Differential Calculus
Differentiable Bounded Convex Real Function is Constant

== Theorem ==
Let $f$ be a real function which is:
:$(1): \quad$ differentiable on $\R$
:$(2): \quad$ bounded on $\R$
:$(3): \quad$ convex on $\R$.
Then $f$ is constant.
== Proof ==
Let $f$ be differentiable and bounded on $\R$.
Let $f$ be convex on $\R$.
Let $\xi \in \R$.
 $\map {f'} \xi > 0$.
Then by Mean Value of Convex Real Function it follows that:
:$\map f x \ge \map f \xi + \map {f'} \xi \paren {x - \xi} \to + \infty$ as $x \to +\infty$
and therefore is not bounded.
Similarly, suppose $\map {f'} \xi < 0$.
Then by Mean Value of Convex Real Function it follows that:
:$\map f x \ge \map f \xi + \map {f'} \xi \paren {x - \xi} \to + \infty$ as $x \to -\infty$
and therefore is likewise not bounded.
Hence $\map {f'} \xi = 0$.
From Zero Derivative implies Constant Function, it follows that $f$ is constant.
== Also see ==
* Differentiable Bounded Concave Real Function is Constant
== Sources ==
* : $\S 12.21 \ (5)$
Category:Convex Real Functions
Category:Differential Calculus
Differentiable Function as Distribution

== Theorem ==
Let $T \in \map {\DD'} \R$ be a distribution.
Let $f : \R \to \R$ be a continuously differentiable real function.
Suppose $T_f$ is a distribution identified with $f$.
Then $T_f' = T_{f'}$.
== Proof ==
Let $\phi \in \map \DD \R$ be a test function with a support on $\closedint a b$.
Then:
== Sources ==
* : Chapter $\S 6.2$: A glimpse of distribution theory. Derivatives in the distributional sense
Category:Examples of Distributions
Category:Functional Analysis
Differentiable Function is Continuous

== Theorem ==
Let $f$ be a real function defined on an interval $I$.
Let $x_0 \in I$ such that $f$ is differentiable at $x_0$.
Then $f$ is continuous at $x_0$.
=== Corollary ===
== Proof ==
We have that $\map {f'} {x_0}$ exists.
Let $x, x_0 \in I$ such that $x \ne x_0$.
Then:
Thus:
:$\map f x \to \map f {x_0}$ as $x \to x_0$
or in other words:
:$\ds \lim_{x \mathop \to x_0} \map f x = \map f {x_0}$
The result follows by definition of continuous.
== Also see ==
* Complex-Differentiable Function is Continuous
== Sources ==
* : $\S 10.6$
* 
* 
* 
* 
Category:Differentiable Real Functions
Category:Continuous Real Functions
Differentiable Function is Continuous/Corollary

== Corollary to Differentiable Function is Continuous ==
Let $f$ be a real function defined on an interval $I$.
If $f$ is not continuous at $x_0$, $f$ is not differentiable at $x_0$.
== Proof ==
From Differentiable Function is Continuous, if $f$ is differentiable at $x_0$, then it is continuous at $x_0$.
By the Rule of Transposition, it follows that if $f$ is not continuous at $x_0$, then it can not be differentiable at $x_0$.
== Sources ==
Category:Differentiable Real Functions
Category:Continuous Real Functions
Differentiable Function of Bounded Variation may not have Bounded Derivative

== Theorem ==
Let $a, b$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a continuous function of bounded variation.
Let $f$ be differentiable on $\openint a b$.
Then $f'$ is not necessarily bounded.
== Proof ==
Proof by Counterexample:
Take $a = 0$, $b = 1$. 
Let $f : \closedint 0 1 \to \R$ have: 
:$\map f x = \sqrt x$
for all $x \in \closedint 0 1$. 
Note that $f$ is increasing, so by Monotone Function is of Bounded Variation:
:$f$ is of bounded variation.
By Derivative of Power, $f$ is differentiable on $\openint 0 1$ with derivative:
:$\map {f'} x = \dfrac 1 {2 \sqrt x}$
Note that this is unbounded as $x \to 0^+$.
So $f$ does not have a bounded derivative, despite being of bounded variation.
== Also see ==
* Differentiable Function with Bounded Derivative is of Bounded Variation, of which this is the (false) converse.
Category:Bounded Variation
Differentiable Function with Bounded Derivative is Absolutely Continuous

== Theorem ==
Let $a, b$ be real numbers with $a < b$.
Let $f: \closedint a b \to \R$ be a continuous function.
Let $f$ be differentiable on $\openint a b$, with bounded derivative.
Then $f$ is absolutely continuous.
== Proof ==
Since the derivative of $f$ is bounded, there exists some $M \in \R_{> 0}$ such that: 
:$\size {\map {f'} x} \le M$
for all $x \in \openint a b$.
Let $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq \closedint a b$ be a collection of disjoint closed real intervals.
Note that for each $i \in \set {1, 2, \ldots, n}$:
:$f$ is continuous on $\closedint {a_i} {b_i}$ and differentiable on $\openint {a_i} {b_i}$.
So, by the Mean Value Theorem, for each $i$ there exists some $\xi_i \in \openint {a_i} {b_i}$ such that:
:$\map {f'} {\xi_i} = \dfrac {\map f {b_i} - \map f {a_i} } {b_i - a_i}$
We then have:
Let $\epsilon$ be a positive real number. 
Then for all collections of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq \closedint a b$ with:
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \frac \epsilon M$
we have:
:$\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \epsilon$
Since $\epsilon$ was arbitrary:
:$f$ is absolutely continuous.
Category:Differentiable Real Functions
Category:Absolutely Continuous Real Functions
Differentiable Function with Bounded Derivative is of Bounded Variation

== Theorem ==
Let $a, b$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a continuous function.
Let $f$ be differentiable on $\openint a b$, with bounded derivative.
Then $f$ is of bounded variation.
== Proof ==
For each finite subdivision $P$ of $\closedint a b$, write: 
:$P = \set {x_0, x_1, \ldots, x_n }$
with:
:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
Since the derivative of $f$ is bounded, there exists some $M \in \R$ such that: 
:$\size {\map {f'} x} \le M$
for all $x \in \openint a b$.
By the Mean Value Theorem, for each $i \in \N$ with $i \le n$, there exists $\xi_i \in \openint {x_{i - 1} } {x_i}$ such that: 
:$\map {f'} {\xi_i} = \dfrac {\map f {x_i} - \map f {x_{i - 1} } } {x_i - x_{i - 1} }$
Note that, from the boundedness of $f'$: 
:$\size {\map {f'} {\xi_i} } \le M$
We also have from the fact that $x_i > x_{i - 1}$: 
:$\size {x_i - x_{i - 1} } = x_i - x_{i - 1}$
So, for each $i$: 
We therefore have: 
for all finite subdivisions $P$.
So $f$ is of bounded variation. 
== Also see ==
* Differentiable Function of Bounded Variation may not have Bounded Derivative: demonstrating that while this theorem gives a simple sufficient condition for a differentiable continuous function $f$ to be of bounded variation, it is not a necessary condition.
== Sources ==
* : $\S 6.3$: Functions of Bounded Variation: Theorem $6.6$
Category:Bounded Variation
Differentiable Operator-Valued Function is Continuous

== Theorem ==
Let $\struct {X, \norm \cdot_X}$ be a normed vector space.
Let $f : I \to X$ be a map defined on an interval $I$.
Let $x_0 \in I$ such that $f$ is differentiable at $x_0$.
Then $f$ is continuous at $x_0$.
== Proof ==
We have that the derivative $\map {f'} {x_0}$ of $f$ at $x_0$ exists.
Hence:
Thus:
:$\map f x \to \map f {x_0}$ as $x \to x_0$
The result follows by the definition of continuity in metric spaces and Metric Induced by Norm is Metric.
== Sources ==
* : Chapter $\text {III}$: Vector-Valued Functions: $3.2$. Some properties of vector-valued functions. Before Theorem $3.2.1$
Category:Bounded Linear Transformations
Differential Entropy of Continuous Uniform Distribution

== Theorem == 
Let $X \sim \ContinuousUniform a b$ for some $a, b \in \R$, $a \ne b$, where $\operatorname U$ is the continuous uniform distribution. 
Then the differential entropy of $X$, $\map h X$, is given by: 
:$\map h X = \map \ln {b - a}$
== Proof == 
From the definition of the continuous uniform distribution, $X$ has probability density function: 
:$\map {f_X} x = \begin{cases} \dfrac 1 {b - a} & : a \le x \le b \\ 0 & : \text{otherwise} \end{cases}$
From the definition of differential entropy: 
:$\ds \map h X = - \int_{-\infty}^\infty \map {f_X} x \map \ln {\map {f_X} x} \rd x$
So: 
Category:Continuous Uniform Distribution
Category:Differential Entropy
Differential Entropy of Exponential Distribution

== Theorem ==
Let $X$ be a continuous random variable of the exponential distribution with parameter $\beta$ for some $\beta \in \R_{> 0}$. 
Then the differential entropy of $X$, $\map h X$, is given by: 
:$\map h X = 1 + \map \ln \beta$
== Proof ==
From the definition of the exponential distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 \beta e^{-\frac x \beta}$
From the definition of differential entropy: 
:$\ds \map h X = -\int_0^\infty \map {f_X} x \map \ln {\map {f_X} x} \rd x$
So:
Category:Differential Entropy
Category:Exponential Distribution
Differential Entropy of Normal Distribution

== Theorem == 
Let $X \sim \Gaussian \mu {\sigma^2}$ for some $\mu \in \R, \sigma \in \R_{> 0}$, where $N$ is the normal distribution.
Then the differential entropy $\map h X$ of $X$ is given by: 
:$\map h X = \map \ln {\sigma \sqrt {2 \pi} } + \dfrac 1 2$
== Proof == 
From the definition of the normal distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 {\sigma \sqrt {2 \pi} } \, \map \exp {-\dfrac {\paren {x - \mu}^2} {2 \sigma^2} }$
From the definition of differential entropy: 
:$\ds \map h X = -\int_{-\infty}^\infty \map {f_X} x \ln \map {f_X} x \rd x$
So:
Category:Normal Distribution
Category:Differential Entropy
Differential Equation/Examples

== Examples of Differential Equations ==
=== First Order Linear Ordinary Differential Equation ===
=== Second Order Linear Ordinary Differential Equation ===
=== First Order First Degree Non-Linear Ordinary Differential Equation $(1)$ ===
=== First Order First Degree Non-Linear Ordinary Differential Equation $(2)$ ===
=== First Order First Degree Non-Linear Ordinary Differential Equation $(3)$ ===
=== First Order Second Degree Non-Linear Ordinary Differential Equation ===
=== Second Order Non-Linear Ordinary Differential Equation ===
=== Second Order Second Degree Non-Linear Ordinary Differential Equation ===
=== Third Order Linear Ordinary Differential Equation ===
=== Third Order Second Degree Non-Linear Ordinary Differential Equation ===
=== Fourth Order First Degree Non-Linear Ordinary Differential Equation ===
=== First Order Linear Partial Differential Equation ===
=== Second Order Linear Partial Differential Equation $(1)$ ===
=== Second Order Linear Partial Differential Equation $(2)$ ===
=== Second Order Second Degree Non-Linear Partial Differential Equation ===
=== First Order First Degree Total Differential Equation ===
=== First Order Second Degree Total Differential Equation ===
Category:Examples of Differential Equations
Differential Equation/Examples/First Order First Degree Non-Linear Ordinary 1

== Example of Differential Equation ==
Ordinary non-linear differential equation of the $1$st order and $1$st degree:
:$\paren {x + y}^2 \dfrac {\d y} {\d x} = 1$
== Sources ==
* : Chapter $\text I$: Introductory: $\S 1.1$ Definitions
Category:Examples of Differential Equations
Differential Equation/Examples/First Order First Degree Non-Linear Ordinary 2

== Example of Differential Equation ==
Ordinary non-linear differential equation of the $1$st order and $1$st degree:
:$\dfrac {\d y} {\d x} = \dfrac x {y^{1/2} \paren {1 + x^{1/2} } }$
== Sources ==
* : Chapter $\text I$: Introduction and Definitions. Elimination. Graphical Representation: $1 \ (4)$
Category:Examples of Differential Equations
Differential Equation/Examples/First Order First Degree Non-Linear Ordinary 3

== Example of Differential Equation ==
Ordinary non-linear differential equation of the $1$st order and $1$st degree:
:$\dfrac {\d y} {\d x} = 1 + x y^2$
== Sources ==
* : Chapter $\text I$: Existence of Solutions: $1$. Some problems for investigation
Category:Examples of Differential Equations
Differential Equation/Examples/First Order Linear Partial

== Example of Differential Equation ==
Partial linear differential equation of the $1$st order in $2$ independent variables:
:$x \dfrac {\partial z} {\partial x} + y \dfrac {\partial z} {\partial y} - z = 0$
== Sources ==
* : Chapter $\text I$: Introductory: $\S 1.1$ Definitions
Category:Examples of Differential Equations
Differential Equation/Examples/First Order Second Degree Total

== Example of Differential Equation ==
Total differential equation of the $1$st order and $2$nd degree:
:$x^2 \rd x^2 + 2 x y \rd x \rd y + y^2 \rd y^2 - z^2 \rd z^2 = 0$
== Sources ==
* : Chapter $\text I$: Introductory: $\S 1.1$ Definitions
Category:Examples of Differential Equations
Differential Equation/Examples/Fourth Order First Degree Non-Linear Ordinary

== Example of Differential Equation ==
Ordinary non-linear differential equation of the $4$th order and $1$st degree:
:$x \dfrac {\d^4 y} {\d x^4} + 2 \dfrac {\d^2 y} {\d x^2} + \paren {x \dfrac {\d y} {\d x} }^5 = x^3$
== Proof ==
The term in $\dfrac {\d y} {\d x}$ is raised to the $5$th degree and so the equation is non-linear.
The term in $\dfrac {\d^4 y} {\d x^4}$ is raised to the $1$st degree.
The result follows by definition of degree.
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation: $(3.18)$
Category:Examples of Differential Equations
Differential Equation/Examples/Second Order Linear Ordinary

== Example of Differential Equation ==
Ordinary linear differential equation of the $2$nd order:
:$\dfrac {\d^2 y} {\d x^2} + y = x^2$
== Sources ==
* : Chapter $\text I$: Introductory: $\S 1.1$ Definitions
Category:Examples of Differential Equations
Differential Equation/Examples/Second Order Linear Partial 1

== Example of Differential Equation ==
Partial linear differential equation of the $2$nd order in $3$ independent variables:
:$\dfrac {\partial^2 V} {\partial x^2} + \dfrac {\partial^2 V} {\partial y^2} + \dfrac {\partial^2 V} {\partial z^2} = 0$
== Sources ==
* : Chapter $\text I$: Introductory: $\S 1.1$ Definitions
Category:Examples of Differential Equations
Differential Equation/Examples/Second Order Linear Partial 2

== Example of Differential Equation ==
Partial linear differential equation of the $2$nd order in $2$ independent variables:
:$\dfrac {\partial^2 y} {\partial t^2} = \alpha^2 \dfrac {\partial^2 y} {\partial x^2}$
== Sources ==
* : Chapter $\text I$: Introduction and Definitions. Elimination. Graphical Representation: $1 \ (5)$
Category:Examples of Differential Equations
Differential Equation/Examples/Second Order Non-Linear Ordinary

== Example of Differential Equation ==
Ordinary non-linear differential equation of the $2$nd order:
:$\dfrac {\d^2 y} {\d x^2} + \paren {3 \dfrac {\d y} {\d x} }^3 + 2 x = 7$
== Proof ==
The term in $\dfrac {\d y} {\d x}$ is raised to the $3$rd degree and so the equation is non-linear.
The term in $\dfrac {\d^2 y} {\d x^2}$ is to the $1$st degree.
Hence the result by definition of degree.
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation: $(3.16)$
Category:Examples of Differential Equations