nyuuzyou/wikis huggingface dataset:
Linear Combination of Riemann-Stieltjes Integrals/Integrator

== Theorem ==
Let $f, \alpha, \beta$ be real functions that are bounded on $\closedint a b$.
Let $c_1, c_2 \in \R$.
Suppose that $f$ is Riemann-Stieltjes integrable with respect both $\alpha$ and $\beta$ on $\closedint a b$ and:
:$\ds \int_a^b f \rd \alpha = A$
:$\ds \int_a^b f \rd \beta = B$
Then, $f$ is Riemann-Stieltjes integrable with respect to $\gamma : \closedint a b \to \R$ defined as:
:$\map \gamma x = c_1 \map \alpha x + c_2 \map \beta x$
and furthermore:
:$\ds \int_a^b f \rd \gamma = c_1 A + c_2 B$
== Proof ==
Let $\epsilon > 0$ be arbitrary.
By definition of the Riemann-Stieltjes integral, let $P'_\epsilon, P' '_\epsilon$ be subdivisions of $\closedint a b$ such that:
:For every $P$ finer than $P'_\epsilon$, $\size {\map S {P, f, \alpha} - A} < \dfrac \epsilon {\size {c_1} + \size {c_2} + 1}$
:For every $P$ finer than $P' '_\epsilon$, $\size {\map S {P, f, \beta} - B} < \dfrac \epsilon {\size {c_1} + \size {c_2} + 1}$
Define $P_\epsilon = P'_\epsilon \cup P' '_\epsilon$ to be a subdivision that is finer than both $P'_\epsilon$ and $P' '_\epsilon$.
Let $P = {x_0, \dotsc, x_n} \supseteq P_\epsilon$ be an arbitrary subdivision that is finer than $P_\epsilon$.
Clearly, $P$ is also finer than $P'_\epsilon$ and $P' '_\epsilon$.
Finally, fix the $t_1, \dotsc, t_k$ for the following Riemann-Stieltjes sum.
We have:
And so:
As $\epsilon > 0$ was arbitrary, by the definition of the Riemann-Stieltjes integral:
:$\ds \int_a^b f \rd \gamma = c_1 A + c_2 B$
== Sources ==
* : Chapter $7$ The Riemann-Stieltjes Integral: $\S 7.4$: Theorem $7.3$
Category:Riemann-Stieltjes Integral
Category:Integral Calculus
Linear Combination of Sequence is Linear Combination of Set

== Theorem ==
Let $G$ be an $R$-module.
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of elements of $G$.
Let $b$ be an element of $G$.
Then:
:$b$ is a linear combination of the sequence $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$
:
:$b$ is a linear combination of the set $\set {a_k: 1 \mathop \le k \mathop \le n}$
== Proof ==
=== Necessary Condition ===
By definition of linear combination of subset:
:Every linear combination of $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ is a linear combination of $\set {a_k: 1 \mathop \le k \mathop \le n}$.
=== Sufficient Condition ===
Let $b$ be a linear combination of $\set {a_k: 1 \mathop \le k \mathop \le n} = \set {a_1, a_2, \ldots, a_n}$.
Then there exists:
:a sequence $\sequence {c_j}_{1 \mathop \le j \mathop \le m}$ of elements of $\set {a_1, a_2, \ldots, a_n}$
and:
:a sequence $\sequence {\mu_j}_{1 \mathop \le j \mathop \le m}$ of scalars such that:
::$\ds b = \sum_{j \mathop = 1}^m \mu_j c_j$
For each $k \in \closedint 1 n$, let $\lambda_k$ be defined as follows.
If:
:$a_k \in \set {c_1, c_2, \ldots, c_m}$
and:
:$a_i \ne a_j$ for all indices $i$ such that $1 \le i < k$
let $\lambda_k$ be the sum of all scalars $\mu_j$ such that $c_j = a_k$.
If:
:$a_k \notin \set {c_1, c_2, \ldots, c_m}$
or:
:$a_i = a_j$ for some index $i$ such that $1 \le i < k$
let $\lambda_k = 0$.
It follows that:
:$\ds b = \sum_{j \mathop = 1}^m \mu_j c_j = \sum_{k \mathop = 1}^n \lambda_k a_k$
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ and $\sequence {b_j}_{1 \mathop \le j \mathop \le m}$ be sequences of elements of $G$ such that $\set {a_1, a_2, \ldots, a_n}$ and $\set {b_1, b_2, \ldots, b_m}$ are identical.
Then as a consequence of the above:
:an element is a linear combination of $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$
:
:it is a linear combination of $\set {a_k: 1 \mathop \le k \mathop \le n}$
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases
Category:Linear Algebra
Linear Combination of Signed Measures is Signed Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ and $\nu$ be signed measures on $\struct {X, \Sigma}$.
Let $\alpha, \beta \in \overline \R$. 
Suppose that the sum: 
:$\alpha \map \mu A + \beta \map \nu A$
is well-defined for each $A \in \Sigma$. 
Then: 
:$\xi = \alpha \mu + \beta \nu$ is a signed measure.
 
== Proof ==
We verify both of the conditions for a signed measure.
=== Proof of $(1)$ ===
We have:
verifying $(1)$ for $\xi$.
=== Proof of $(2)$ ===
Let $\sequence {D_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint $\Sigma$-measurable sets. 
Then: 
verifying $(2)$ for $\xi$.
So $\xi$ is a signed measure.
Category:Signed Measures
Linear Combination of Solutions to Homogeneous Linear 2nd Order ODE

== Theorem ==
Let $c_1$ and $c_2$ be real numbers.
Let $\map {y_1} x$ and $\map {y_2} x$ be particular solutions to the homogeneous linear second order ODE:
:$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
Then:
:$c_1 \, \map {y_1} x + c_2 \, \map {y_2} x$
is also a particular solution to $(1)$.
That is, a linear combination of particular solutions to a homogeneous linear second order ODE is also a particular solution to that ODE.
== Proof ==
Hence the result.
== Sources ==
* : $\S 3.14$: Second Order Linear Equations: Introduction: Theorem $\text{C}$
Category:Homogeneous LSOODEs
Linear Combination of Weakly Convergent Sequences is Weakly Convergent

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\Bbb F$.
Let $x, y \in X$.
Let $\alpha, \beta \in \Bbb F$. 
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$ converging weakly to $x$. 
Let $\sequence {y_n}_{n \mathop \in \N}$ be a sequence in $X$ converging weakly to $y$. 
Then: 
:$\alpha x_n + \beta y_n \weakconv \alpha x + \beta y$
where $\weakconv$ denotes weak convergence.
== Proof ==
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $\struct {X, \norm \cdot}$. 
Since: 
:$x_n \weakconv x$
We have: 
:$\map f {x_n} \to \map f x$ 
for each $f \in X^\ast$.
Similarly, since: 
:$y_n \weakconv y$
we have: 
:$\map f {y_n} \to \map f y$ 
From Combined Sum Rule for Real Sequences, we therefore have: 
:$\alpha \map f {x_n} + \beta \map f {y_n} \to \alpha \map f x + \beta \map f {y_n}$
Since $f$ is linear, we have: 
:$\alpha \map f {x_n} + \beta \map f {y_n} = \map f {\alpha x_n + \beta y_n}$ for each $n \in \N$.
and:
:$\alpha \map f x + \beta \map f y = \map f {\alpha x + \beta y}$
So we have: 
:$\map f {\alpha x_n + \beta y_n} \to \map f {\alpha x + \beta y}$
for each $f \in X^\ast$. 
So, from the definition of weak convergence, we have: 
:$\alpha x_n + \beta y_n \weakconv \alpha x + \beta y$
Category:Weak Convergence (Normed Vector Spaces)
Linear Congruence/Examples/15 x equiv 6 mod 18

== Example of Linear Congruence ==
The linear congruence:
:$15 x \equiv 6 \pmod {18}$
has $3$ solutions:
:$x = 4, 10, 16$
== Proof ==
We have that:
:$\gcd \set {15, 18} = 3$
Then we note that $3$ is a divisor of $6$.
Hence by Solution of Linear Congruence, $15 x \equiv 6 \pmod {18}$ has $3$ solutions in the integers, given by:
:$\dfrac {15} 3 x \equiv \dfrac 6 3 \paren {\bmod \dfrac {18} 3}$
that is:
:$5 x \equiv 2 \pmod 6$
which leads us to:
:$x = 4 + 2 t$
for $t = 0, 1, 2$.
Hence the result.
== Sources ==
* 
* 
Category:Examples of Linear Congruences
Linear Congruence/Examples/2 x equiv 7 mod 18

== Example of Linear Congruence ==
The linear congruence:
:$2 x \equiv 7 \pmod {18}$
has no solution in $\Z$.
== Proof ==
We have that:
:$\gcd \set {2, 18} = 2$
But $2$ is not a divisor of $7$.
Hence by Solution of Linear Congruence, $2 x \equiv 7 \pmod {18}$ has no solutions in the integers.
== Sources ==
* 
* 
Category:Examples of Linear Congruences
Linear Congruence/Examples/7 x equiv 8 mod 30

== Example of Linear Congruence ==
The linear congruence:
:$7 x \equiv 8 \pmod {30}$
has the unique solution:
:$x = 14$
== Proof ==
We have that:
:$\gcd \set {7, 30} = 1$
Hence by Solution of Linear Congruence, $7 x \equiv 8 \pmod {30}$ has $1$ solution in the integers, given by:
which, after working, gives us:
:$x = 14$
Hence the result.
== Sources ==
* 
* 
Category:Examples of Linear Congruences
Linear Diophantine Equation/Examples

== Examples of Linear Diophantine Equations ==
=== Example: $15 x + 27 y = 1$ ===
=== Example: $5 x + 6 y = 1$ ===
=== Example: $2 x + 3 y = 4$ ===
=== Example: $17 x + 19 y = 23$ ===
=== Example: $15 x + 51 y = 41$ ===
=== Example: $23 x + 29 y = 25$ ===
=== Example: $10 x - 8 y = 42$ ===
=== Example: $121 x - 88 y = 572$ ===
=== Example: $17 x + 15 y = 143$ ===
=== Example: $35 x - 256 y = 48$ ===
Category:Examples of Linear Diophantine Equations
Linear Diophantine Equation/Examples/10x - 8y = 42

== Example of Linear Diophantine Equation ==
The linear diophantine equation:
:$10 x - 8 y = 42$
has the general solution:
:$\tuple {x, y} = \tuple {21 - 4 t, 21 - 5 t}$
== Proof ==
Using the Euclidean Algorithm:
Thus we have that:
:$\gcd \set {10, -8} = 2$
which is a divisor of $42$:
:$42 = 21 \times 2$
So, from Solution of Linear Diophantine Equation, a solution exists.
Next we find a single solution to $10 x - 8 y = 42$.
Again with the Euclidean Algorithm:
and so:
is a solution.
From Solution of Linear Diophantine Equation, the general solution is:
:$\forall t \in \Z: x = x_0 + \dfrac b d t, y = y_0 - \dfrac a d t$
where $d = \gcd \set {a, b}$.
In this case:
giving:
== Sources ==
* : $\text {2-3}$ The Linear Diophantine Equation: Exercise $1 \ \text {(e)}$
Category:Examples of Linear Diophantine Equations
Linear Diophantine Equation/Examples/121x - 88y = 572

== Example of Linear Diophantine Equation ==
The linear diophantine equation:
:$121 x - 88 y = 572$
has the general solution:
:$\tuple {x, y} = \tuple {156 - 8 t, 208 - 11 t}$
== Proof ==
Using the Euclidean Algorithm:
Thus we have that:
:$\gcd \set {121, -88} = 11$
which is a divisor of $572$:
:$572 = 52 \times 11$
So, from Solution of Linear Diophantine Equation, a solution exists.
Next we find a single solution to $121 x - 88 y = 572$.
Again with the Euclidean Algorithm:
and so:
is a solution.
From Solution of Linear Diophantine Equation, the general solution is:
:$\forall t \in \Z: x = x_0 + \dfrac b d t, y = y_0 - \dfrac a d t$
where $d = \gcd \set {a, b}$.
In this case:
giving:
== Sources ==
* : $\text {2-3}$ The Linear Diophantine Equation: Exercise $1 \ \text {(f)}$
Category:Examples of Linear Diophantine Equations
Linear Diophantine Equation/Examples/15x + 27y = 1

== Example of Linear Diophantine Equation ==
The linear diophantine equation:
:$15 x + 27 y = 1$
has no solutions for $x$ and $y$ integers.
== Proof ==
We have that:
:$\gcd \set {15, 27} = 3$
which is not a divisor of $1$.
So, from Solution of Linear Diophantine Equation, no solution exists.
== Sources ==
* : $\text {2-3}$ The Linear Diophantine Equation: Example $\text {2-12}$
Category:Examples of Linear Diophantine Equations
Linear Diophantine Equation/Examples/15x + 51y = 41

== Example of Linear Diophantine Equation ==
The linear diophantine equation:
:$15 x + 51 y = 41$
has no solutions for $x$ and $y$ integers.
=== Graphical Presentation ===
== Proof ==
Using the Euclidean Algorithm:
Thus we have that:
:$\gcd \set {15, 45} = 3$
which is not a divisor of $41$.
So, from Solution of Linear Diophantine Equation, no solution exists.
== Sources ==
* : $\text {2-3}$ The Linear Diophantine Equation: Exercise $1 \ \text {(c)}$
Category:Examples of Linear Diophantine Equations
Linear Diophantine Equation/Examples/15x + 51y = 41/Graph

== Graphical Presentation of Linear Diophantine Equation: $15 x + 51 y = 41$ ==
:
== Sources ==
* : $\text {2-3}$ The Linear Diophantine Equation: Exercise $3 \ \text{(c)}$
Category:Examples of Linear Diophantine Equations
Linear Diophantine Equation/Examples/17x + 15y = 143

== Example of Linear Diophantine Equation ==
The linear diophantine equation:
:$17 x + 15 y = 143$
has the general solution in positive integers:
:$\tuple {x, y} = \tuple {4, 5}$
== Proof ==
Using the Euclidean Algorithm:
Thus we have that:
:$\gcd \set {17, 15} = 1$
which is (trivially) a divisor of $143$.
So, from Solution of Linear Diophantine Equation, a solution exists.
Next we find a single solution to $17 x + 15 y = 143$.
Again with the Euclidean Algorithm:
and so:
is a solution.
From Solution of Linear Diophantine Equation, the general solution is:
:$\forall t \in \Z: x = x_0 + \dfrac b d t, y = y_0 - \dfrac a d t$
where $d = \gcd \set {a, b}$.
In this case:
giving:
Both $x$ and $y$ must be non-negative integers.
Hence:
and:
Hence $t = 67$ and so:
The required solution is therefore:
:$x = 4, y = 5$
== Historical Note ==
== Sources ==
* : $\text {2-3}$ The Linear Diophantine Equation: Exercise $2$
Category:Examples of Linear Diophantine Equations
Linear Diophantine Equation/Examples/17x + 15y = 143/Historical Note

== Historical Note on Linear Diophantine Equation: $17 x + 15 y = 143$ ==
This exercise was originally couched in the language of a person making a purchase of:
:$x$ items of commodity $A$, costing $17$ monetary units each
and:
:$y$ items of commodity $B$, costing $15$ monetary units each
for a total outlay of $143$ monetary units.
It is apparent in this context that both $x$ and $y$ must be non-negative, and (in the specific wording of the question) can be interpreted as actually strictly positive.
Such exercises are usually (as this one is) specifically crafted so as to allow for only one solution.
== Sources ==
* : $\text {2-3}$ The Linear Diophantine Equation: Exercise $2$
Category:Examples of Linear Diophantine Equations
Linear Diophantine Equation/Examples/17x + 19y = 23

== Example of Linear Diophantine Equation ==
The linear diophantine equation:
:$17 x + 19 y = 23$
has the general solution:
:$\tuple {x, y} = \tuple {207 + 19 t, -184 - 17 t}$
=== Graphical Presentation ===
== Proof ==
Using the Euclidean Algorithm:
Thus we have that:
:$\gcd \set {17, 19} = 1$
which is (trivially) a divisor of $23$.
So, from Solution of Linear Diophantine Equation, a solution exists.
Next we find a single solution to $17 x + 19 y = 23$.
Again with the Euclidean Algorithm:
and so:
is a solution.
From Solution of Linear Diophantine Equation, the general solution is:
:$\forall t \in \Z: x = x_0 + \dfrac b d t, y = y_0 - \dfrac a d t$
where $d = \gcd \set {a, b}$.
In this case:
giving:
== Sources ==
* : $\text {2-3}$ The Linear Diophantine Equation: Exercise $1 \ \text {(b)}$
Category:Examples of Linear Diophantine Equations
Linear Diophantine Equation/Examples/17x + 19y = 23/Graph

== Graphical Presentation of Linear Diophantine Equation: $17 x + 19 y = 23$ ==
:
== Sources ==
* : $\text {2-3}$ The Linear Diophantine Equation: Exercise $3 \ \text{(b)}$
Category:Examples of Linear Diophantine Equations
Linear Diophantine Equation/Examples/23x + 29y = 25

== Example of Linear Diophantine Equation ==
The linear diophantine equation:
:$23 x + 29 y = 25$
has the general solution:
:$\tuple {x, y} = \tuple {-125 + 29 t, 100 - 23 t}$
== Proof ==
Using the Euclidean Algorithm:
Thus we have that:
:$\gcd \set {23, 29} = 1$
which is (trivially) a divisor of $25$.
So, from Solution of Linear Diophantine Equation, a solution exists.
Next we find a single solution to $23 x + 29 y = 25$.
Again with the Euclidean Algorithm:
and so:
is a solution.
From Solution of Linear Diophantine Equation, the general solution is:
:$\forall t \in \Z: x = x_0 + \dfrac b d t, y = y_0 - \dfrac a d t$
where $d = \gcd \set {a, b}$.
In this case:
giving:
== Sources ==
* : $\text {2-3}$ The Linear Diophantine Equation: Exercise $1 \ \text {(d)}$
Category:Examples of Linear Diophantine Equations
Linear Diophantine Equation/Examples/2x + 3y = 4

== Example of Linear Diophantine Equation ==
The linear diophantine equation:
:$2 x + 3 y = 4$
has the general solution:
:$x = -4 + 3 t, y = 4 - 2 t$
=== Graphical Presentation ===
== Proof ==
We have that:
:$\gcd \set {2, 3} = 1$
which is (trivially) a divisor of $4$.
So, from Solution of Linear Diophantine Equation, a solution exists.
First we find a single solution to $2 x + 3 y = 4$:
and so:
is a solution.
From Solution of Linear Diophantine Equation, the general solution is:
:$\forall t \in \Z: x = x_0 + \dfrac b d t, y = y_0 - \dfrac a d t$
where $d = \gcd \set {a, b}$.
In this case:
giving:
== Sources ==
* : $\text {2-3}$ The Linear Diophantine Equation: Exercise $1 \ \text {(a)}$
Category:Examples of Linear Diophantine Equations
Linear Diophantine Equation/Examples/2x + 3y = 4/Graph

== Graphical Presentation of Linear Diophantine Equation: $2 x + 3 y = 4$ ==
:
== Sources ==
* : $\text {2-3}$ The Linear Diophantine Equation: Exercise $3 \ \text{(a)}$
Category:Examples of Linear Diophantine Equations
Linear Diophantine Equation/Examples/35x - 256y = 48

== Example of Linear Diophantine Equation ==
The linear diophantine equation:
:$35 x - 256 y = 48$
has the general solution:
:$\tuple {x, y} = \tuple {16 + 256 t, 2 + 35 t}$
== Proof ==
We use Solution of Linear Diophantine Equation.
Using the Euclidean Algorithm:
Hence we see that $\gcd \set {35, -256} = 1$ which trivially divides $48$, and so there exists a solution.
Again with the Euclidean Algorithm:
From Solution of Linear Diophantine Equation, the general solution is:
:$\tuple {x, y} = \tuple {-5616 + \paren {-256} t, -768 - 35 t}$
for $t \in \Z$.
This can be simplified by setting $t \to -t$, thus
:$\tuple {x, y} = \tuple {-5616 + 256 t, -768 + 35 t}$
We have that:
:$-5616 + 22 \times 256 = 16$
and:
:$-768 + 22 \times 35 = 2$
hence giving us the answer in smallest positive integer $\tuple {x, y}$:
:$\tuple {x, y} = \tuple {16 + 256 t, 2 + 35 t}$
Category:Examples of Linear Diophantine Equations
Linear Diophantine Equation/Examples/5x + 6y = 1

== Example of Linear Diophantine Equation ==
The linear diophantine equation:
:$5 x + 6 y = 1$
has the general solution:
:$x = -1 + 6 t, y = 1 - 5 t$
== Proof ==
We have that:
:$\gcd \set {5, 6} = 1$
which is (trivially) a divisor of $1$.
So, from Solution of Linear Diophantine Equation, a solution exists.
First we find a single solution to $5 x + 6 y = 1$:
:$1 = 1 \times 6 - 1 \times 5$
So $y_0 = 1, x_0 = -1$ is a solution.
From Solution of Linear Diophantine Equation, the general solution is then:
:$\forall t \in \Z: x = x_0 + \dfrac b d t, y = y_0 - \dfrac a d t$
where $d = \gcd \set {a, b}$.
In this case:
giving:
== Sources ==
* : $\text {2-3}$ The Linear Diophantine Equation: Example $\text {2-13}$
Category:Examples of Linear Diophantine Equations
Linear First Order ODE/(1 + x^2) dy + 2 x y dx = cotangent x dx

== Theorem ==
The linear first order ODE:
:$(1): \quad \paren {1 + x^2} \rd y + 2 x y \rd x = \cot x \rd x$
has the general solution:
:$y = \dfrac {\map \ln {\sin x} } {1 + x^2} + \dfrac C {1 + x^2}$
== Proof ==
$(1)$ can be written as:
:$(2): \quad \paren {1 + x^2} \dfrac {\rd y} {\rd x} + 2 x y = \cot x$
We notice straight away that:
:$\dfrac {\rd} {\rd x} \paren {1 + x^2} = 2 x$
and so:
:$\dfrac {\rd} {\rd x} \paren {1 + x^2} y = \cot x$
Thus the general solution becomes:
or:
:$y = \dfrac {\map \ln {\sin x} } {1 + x^2} + \dfrac C {1 + x^2}$
== Sources ==
* : $\S 2.10$: Problem $2 \ \text{(c)}$
Category:Examples of Linear First Order ODEs
Linear First Order ODE/(1 + x^2) y' + 2 x y = 4 x^3

== Theorem ==
The linear first order ODE:
:$\paren {1 + x^2} \dfrac {\d y} {\d x} + 2 x y = 4 x^3$
has the general solution:
:$y = \dfrac {x^4} {1 + x^2} + \dfrac C {1 + x^2}$
== Proof ==
It is noticed that:
:$\dfrac {\d} {\d x} \paren {1 + x^2} = 2 x$
and so $(1)$ can be rendered in the form:
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $21$
Category:Examples of Linear First Order ODEs
Linear First Order ODE/(2 y - x^3) dx = x dy

== Theorem ==
The linear first order ODE:
:$(1): \quad \paren {2 y - x^3} \rd x = x \rd y$
has the general solution:
:$y = -x^3 + C x^2$
== Proof ==
Rearranging $(1)$:
:$(2): \quad \dfrac {\d y} {\d x} - \dfrac 2 x y = - x^2$
$(2)$ is in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where $\map P x = -\dfrac 2 x$.
Thus:
Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
:$\map {\dfrac \d {\d x} } {\dfrac y {x^2} } = -1$
and the general solution is:
:$\dfrac y {x^2} = -x + C$
or:
:$y = -x^3 + C x^2$
== Sources ==
* : $\S 2.10$: Problem $2 \ \text{(f)}$
Category:Examples of Linear First Order ODEs
Linear First Order ODE/(x^2 + y) dx = x dy

== Theorem ==
The linear first order ODE:
:$(1): \quad \paren {x^2 + y} \rd x = x \rd y$
has the general solution:
:$y = x^2 + C x$
== Proof ==
Rearranging $(1)$:
:$(2): \quad \dfrac {\d y} {\d x} - \dfrac y x = x$
$(2)$ is in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where:
:$\map P x = -\dfrac 1 x$
:$\map Q x = x$
Thus:
Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
:$\map {\dfrac \d {\d x} } {\dfrac y x} = 1$
and the general solution is:
:$\dfrac y x = x + C$
or:
:$y = x^2 + C x$
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $12$
Category:Examples of Linear First Order ODEs
Linear First Order ODE/dy = f(x) dx

== Theorem ==
Let $f: \R \to \R$ be an integrable real function.
The linear first order ODE:
:$(1): \quad \dfrac {\d y} {\d x} = \map f x$
has the general solution:
:$y = \ds \int \map f x \rd x + C$
where $\ds \int \map f x \rd x$ denotes the primitive of $f$.
=== Initial Condition ===
== Proof ==
== Examples ==
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 1$. The first order equation: $\S 1.1$ Introduction: $(2)$
Category:Examples of Linear First Order ODEs
Linear First Order ODE/dy = f(x) dx/Examples/y' = e^-x^2

== Example of Linear First Order ODE: $\d y = \map f x \rd x$ ==
The linear first order ODE:
:$(1): \quad \dfrac {\d y} {\d x} = e^{-x^2}$
has the general solution:
:$y = \dfrac {\sqrt \pi} 2 \map {\erf} x + C$
where $\erf$ denotes the error function.
== Proof ==
From the definition of the error function:
:$\map {\erf} x = \ds \dfrac 2 {\sqrt \pi} \int_0^x \map \exp {-t^2} \rd t$
where $\exp$ is the real exponential function.
So by the Fundamental Theorem of Calculus:
:$\dfrac {\rd} {\rd x} \map {\erf} x = \dfrac 2 {\sqrt \pi} e^{-x^2} + C$
The result follows.
== Sources ==
* : Chapter $1$ First-Order Differential Equations: $2$ Fundamental Theorem of the Calculus
Category:Examples of Linear First Order ODEs
Category:Error Function
Linear First Order ODE/dy = f(x) dx/Initial Condition

== Theorem ==
Let $f: \R \to \R$ be an integrable real function.
Consider the linear first order ODE:
:$(1): \quad \dfrac {\d y} {\d x} = \map f x$
subject to the initial condition:
:$y = y_0$ when $x = x_0$
$(1)$ has the particular solution:
:$y = y_0 + \ds \int_{x_0}^x \map f \xi \rd \xi$
where $\ds \int \map f x \rd x$ denotes the primitive of $f$.
== Proof ==
It is seen that $(1)$ is an instance of the first order ordinary differential equation:
:$\dfrac {\d y} {\d x} = \map f {x, y}$
which is:
:subject to an initial condition: $\tuple {x_0, y_0}$
where:
:$\map f {x, y}$ is actually $\map f x$
From Solution to First Order Initial Value Problem, this problem is equivalent to the integral equation:
:$\ds y = y_0 + \int_{x_0}^x \map f {\xi, \map y \xi} \rd \xi$
As $\map y \xi$ does not contribute towards $\map f x$, it can be ignored.
Hence we have:
:$\ds y = y_0 + \int_{x_0}^x \map f \xi \rd \xi$
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 1$. The first order equation: $\S 1.1$ Introduction: $(4)$
Category:Examples of Linear First Order ODEs
Linear First Order ODE/x dy + y dx = x cosine x dx

== Theorem ==
The linear first order ODE:
:$x \rd y + y \rd x = x \cos x \rd x$
has the general solution:
:$x y = x \sin x + \cos x + C$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $8$
Category:Examples of Linear First Order ODEs
Category:Linear First Order ODE/x dy + y dx = x cosine x dx
Linear First Order ODE/x dy + y dx = x cosine x dx/Proof 1

== Theorem ==
== Proof ==
Rearranging:
:$\dfrac {d y} {\d x} + \dfrac y x = \cos x$
This is a linear first order ODE in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where:
:$\map P x y = \dfrac 1 x$
:$\map Q x = \cos x$
Thus:
Thus from Solution by Integrating Factor:
Category:Linear First Order ODE/x dy + y dx = x cosine x dx
Category:Examples of Solution by Integrating Factor