nyuuzyou/wikis huggingface dataset:
First Order ODE/(y - x^3) dx + (x + y^3) dy = 0

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad \paren {y - x^3} \rd x + \paren {x + y^3} \rd y = 0$
is an exact differential equation with solution:
:$4 x y - x^4 + y^4 = C$
This can also be presented as:
:$\dfrac {\d y} {\d x} + \dfrac {y - x^3} {x + y^3} = 0$
== Proof ==
Let:
:$\map M {x, y} = y - x^3$
:$\map N {x, y} = x + y^3$
Then:
Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be exact.
By Solution to Exact Differential Equation, the solution to $(1)$ is:
:$\map f {x, y} = C$
where:
Hence:
and:
Thus:
:$\map f {x, y} = x y - \dfrac {x^4} 4 + \dfrac {y^4} 4$
and by Solution to Exact Differential Equation, the solution to $(1)$ is:
:$x y - \dfrac {x^4} 4 + \dfrac {y^4} 4 = C_1$
which can be simplified by multiplying through by $4$ and setting $C = 4 C_1$:
:$4 x y - x^4 + y^4 = C$
== Sources ==
* : $\S 2.8$: Exact Equations: Problem $3$
Category:Examples of Exact Differential Equation
Category:Examples of First Order ODEs
First Order ODE/(y over x^2) dx + (y - 1 over x) dy = 0

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad \dfrac y {x^2} \rd x + \paren {y - \dfrac 1 x} \rd y = 0$
is an exact differential equation with solution:
:$\dfrac {y^2} 2 - \dfrac y x = C$
== Proof ==
Let $M$ and $N$ be defined as:
:$\map M {x, y} = \dfrac y {x^2}$
:$\map N {x, y} = y - \dfrac 1 x$
Then:
Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be exact.
By Solution to Exact Differential Equation, the solution to $(1)$ is:
:$\map f {x, y} = C$
where:
Hence:
and:
Thus:
:$\map f {x, y} = \dfrac {y^2} 2 - \dfrac y x$
By Solution to Exact Differential Equation, the solution to $(1)$ is:
:$\dfrac {y^2} 2 - \dfrac y x = C$
</onlyinclude>
Category:Examples of Exact Differential Equation
Category:Examples of First Order ODEs
First Order ODE/-1 over y sine x over y dx + x over y^2 sine x over y dy

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad -\dfrac 1 y \sin \dfrac x y \rd x + \dfrac x {y^2} \sin \dfrac x y \rd y = 0$
is an exact differential equation with solution:
:$\dfrac x y = C$
This can also be presented as:
:$\dfrac {\d y} {\d x} = \dfrac {\dfrac 1 y \sin \dfrac x y} {\dfrac x {y^2} \sin \dfrac x y}$
== Proof ==
Let:
:$\map M {x, y} = -\dfrac 1 y \sin \dfrac x y$
:$\map N {x, y} = \dfrac x {y^2} \sin \dfrac x y$
Then:
Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be exact.
By Solution to Exact Differential Equation, the solution to $(1)$ is:
:$\map f {x, y} = C$
where:
Hence:
and:
Set $z = \dfrac x y$:
Thus by Integration by Substitution:
Thus:
:$\map f {x, y} = -\cos \dfrac x y$
and by Solution to Exact Differential Equation, the solution to $(1)$, after simplification, is:
== Sources ==
* : $\S 2.8$: Exact Equations: Problem $8$
Category:Examples of Exact Differential Equation
Category:Examples of First Order ODEs
First Order ODE/1 over x^3 y^2 dx + (1 over x^2 y^3 + 3 y) dy = 0

== Theorem ==
The first order ODE:
:$(1): \quad \dfrac 1 {x^3 y^2} \rd x + \paren {\dfrac 1 {x^2 y^3} + 3 y} \rd y = 0$
is an exact differential equation with solution:
:$-\dfrac 1 {2 x^2 y^2} + \dfrac {3 y^2} 2 = C$
== Proof ==
Let $M$ and $N$ be defined as:
:$\map M {x, y} = \dfrac 1 {x^3 y^2}$
:$\map N {x, y} = \dfrac 1 {x^2 y^3} + 3 y$
Then:
By Solution to Exact Differential Equation, the solution to $(1)$ is:
:$\map f {x, y} = C$
where:
Hence:
and:
Thus:
:$\map f {x, y} = -\dfrac 2 {x^2 y^2} + \dfrac {3 y^2} 2$
and by Solution to Exact Differential Equation, the solution to $(1)$ is:
:$-\dfrac 1 {2 x^2 y^2} + \dfrac {3 y^2} 2 = C$
Category:Examples of Exact Differential Equation
Category:Examples of First Order ODEs
First Order ODE/Cosine (x + y) dx = sine (x + y) dx + x sine (x + y) dy

== Theorem ==
The first order ordinary differential equation:
:$\map \cos {x + y} \rd x = \map \sin {x + y} \rd x + x \map \sin {x + y} \rd y$
is an exact differential equation with solution:
:$x \map \cos {x + y} = C$
== Proof ==
Express it in the form:
:$\paren {x \map \sin {x + y} - \map \cos {x + y} } \rd x + x \map \sin {x + y} \rd y$
Let:
:$\map M {x, y} = x \map \sin {x + y} - \map \cos {x + y}$
:$\map N {x, y} = x \map \sin {x + y}$
Then:
Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be exact.
By Solution to Exact Differential Equation, the solution to $(1)$ is:
:$\map f {x, y} = C$
where:
Hence:
and:
Thus:
:$\map f {x, y} = -x \map \cos {x + y}$
and by Solution to Exact Differential Equation, the solution to $(1)$ is:
:$-x \map \cos {x + y} = C_1$
or setting $C = -C_1$:
:$x \map \cos {x + y} = C$
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $15$
Category:Examples of Exact Differential Equation
Category:Examples of First Order ODEs
First Order ODE/dx = (y over (1 - x^2 y^2)) dx + (x over (1 - x^2 y^2)) dy

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad \d x = \dfrac y {1 - x^2 y^2} \rd x + \dfrac x {1 - x^2 y^2} \rd y$
is an exact differential equation with solution:
:$\map \ln {\dfrac {1 + x y} {1 - x y} } - 2 x = C$
The of this is defined on the set $S$:
:$S = \set {x, y \in \R: \size {x y} < 1}$
where $x, y \in \R$ such that $\size {x y} < 1$.
== Also presented as ==
:$\dfrac {\d y} {\d x} = -\dfrac {\dfrac y {1 - x^2 y^2} - 1} {\dfrac x {1 - x^2 y^2} }$
== Proof ==
First express $(1)$ in the form:
:$(2): \quad \paren {\dfrac y {1 - x^2 y^2} - 1} + \paren {\dfrac x {1 - x^2 y^2} } \dfrac {\d y} {\d x}$
Let:
:$\map M {x, y} = \dfrac y {1 - x^2 y^2} - 1$
:$\map N {x, y} = \dfrac x {1 - x^2 y^2}$
Then:
Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be exact.
By Solution to Exact Differential Equation, the solution to $(1)$ is:
:$\map f {x, y} = C$
where:
Hence:
Substitute $z = x y$ to obtain:
:$\dfrac {\d z} {\d x} = y$
This gives:
and:
Thus:
:$\map f {x, y} = \dfrac 1 2 \map \ln {\dfrac {1 + x y} {1 - x y} } - x$
and by Solution to Exact Differential Equation, the solution to $(1)$, after simplification, is:
:$\map \ln {\dfrac {1 + x y} {1 - x y} } - 2 x = C$
Note that from the , $\map \ln z$ is only defined when $z > 0$.
This means $\dfrac {1 + x y} {1 - x y} > 0$
First assume $x y > 0$:
:$\ds \dfrac {1 + x y} {1 - x y} > 0 \to \paren {1 + x y} > 0 \to x y > -1$
Next assume $x y < 0$:
:$\ds \dfrac {1 + -\paren {x y} } {1 - -\paren {x y} } > 0 \to \paren {1 - x y} > 0 \to x y < 1$
Therefore, $\size {x y} < 1$
== Sources ==
* : $\S 2.8$: Exact Equations: Problem $11$
Category:Examples of Exact Differential Equation
Category:Examples of First Order ODEs
First Order ODE/dy = k y dx

== Theorem ==
Let $k \in \R$ be a real number.
The first order ODE:
:$\dfrac {\d y} {\d x} = k y$
has the general solution:
:$y = C e^{k x}$
== Proof 1 ==
 
== Proof 2 ==
== Proof 3 ==
 
== Also represented as ==
This first order ODE can also be represented as:
:$\dfrac {\d y} {\d x} + k y = 0$
from which the general solution is:
:$y = C e^{-k x}$
== Also see ==
* Decay Equation
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 1$. The first order equation: $\S 1.1$ Introduction: $(5)$
* : $1$: The Nature of Differential Equations: $\S 1$: Introduction: $(4)$
Category:First Order ODE/dy = k y dx
Category:Examples of Linear First Order ODEs
Category:Examples of Solutions to Separable Differential Equation
First Order ODE/dy = k y dx/Proof 2

== Theorem == 
== Proof ==
Write the differential equation as: 
:$y' - k y = 0$
Taking Laplace transforms:
:$\laptrans {y' - k y} = \laptrans 0$
From Laplace Transform of Constant Mapping, we have: 
:$\laptrans 0 = 0$
We also have: 
So:
:$\paren {s - k} \laptrans y = \map y 0$
Giving: 
:$\laptrans y = \dfrac {\map y 0} {s - k}$
So:
Setting $C = \map y 0$ gives the result.
Category:First Order ODE/dy = k y dx
First Order ODE/exp x (1 + x) dx = (x exp x - y exp y) dy

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad e^x \paren {1 + x} \rd x = \paren {x e^x - y e^y} \rd y$
has the solution:
:$2 x e^x e^{-y} + y^2 + C$
== Proof ==
Dividing $(1)$ through by $e^y$:
:$(2): \quad \paren {1 + x} e^x e^{-y} \rd x = \paren {x e^x e^{-y} - y} \rd y$
Let $z = x e^x e^{-y}$.
Then:
:$\dfrac {\d z} {\d x} = \paren {1 + x} e^x e^{-y} - x e^x e^{-y} \dfrac {\d y} {\d x}$
or:
:$\d z = \paren {1 + x} e^x e^{-y} \rd x - x e^x e^{-y} \rd y$
Substituting $z$ and $\d z$ in $(2)$:
:$\d z = \paren {z - y} \rd y - z \rd y = - y \rd y$
which directly leads to:
:$\ds \int \rd z = -\int y \rd y$
from which:
:$z = -\dfrac {y^2} 2 + k$
Substituting for $z$ and letting $C = 2 k$:
:$2 x e^x e^{-y} + y^2 + C$
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $25$
Category:Examples of First Order ODEs
First Order ODE/exp x sine y dx + exp x cos y dy = y sine x y dx + x sine x y dy

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad e^x \sin y \rd x + e^x \cos y \rd y = y \sin x y \rd x + x \sin x y \rd y$
is an exact differential equation with solution:
:$e^x \sin y + \cos x y = C$
== Proof ==
Let $(1)$ be expressed as:
:$\paren {e^x \sin y - y \sin x y} \rd x + \paren {e^x \cos y - x \sin x y} \rd y = 0$
Let:
:$\map M {x, y} = e^x \sin y - y \sin x y$
:$\map N {x, y} = e^x \cos y - x \sin x y$
Then:
Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be exact.
By Solution to Exact Differential Equation, the solution to $(1)$ is:
:$\map f {x, y} = C$
where:
Hence:
and:
Thus:
:$\map f {x, y} = e^x \sin y + \cos x y$
and by Solution to Exact Differential Equation, the solution to $(1)$ is:
:$e^x \sin y + \cos x y = C$
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $22$
Category:Examples of Exact Differential Equation
Category:Examples of First Order ODEs
First Order ODE/exp y dx + (x exp y + 2 y) dy = 0

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad e^y + \paren {x e^y + 2 y} \dfrac {\d y} {\d x} = 0$
is an exact differential equation with solution:
:$x e^y + y^2 = C$
This can also be presented as:
:$\dfrac {\d y} {\d x} + \dfrac {e^y} {x e^y + 2 y} = 0$
== Proof ==
Let:
:$\map M {x, y} = e^y$
:$\map N {x, y} = x e^y + 2 y$
Then:
Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be exact.
By Solution to Exact Differential Equation, the solution to $(1)$ is:
:$\map f {x, y} = C$
where:
Hence:
and:
Thus:
:$\map f {x, y} = x e^y + y^2$
and by Solution to Exact Differential Equation, the solution to $(1)$ is:
:$x e^y + y^2 = C$
== Sources ==
* : $\S 2.8$: Exact Equations: Example $1$
Category:Examples of Exact Differential Equation
Category:Examples of First Order ODEs
First Order ODE/x^2 y' - 3 x y - 2 y^2 = 0

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad x^2 \dfrac {\d y} {\d x} - 3 x y - 2 y^2 = 0$
is a homogeneous differential equation with solution:
:$y = C x^2 \paren {x + y}$
== Proof ==
Let:
:$\map M {x, y} = 3 x y + 2 y^2$
:$\map N {x, y} = x^2$
Put $t x, t y$ for $x, y$:
Thus both $M$ and $N$ are homogeneous functions of degree $2$.
Thus, by definition, $(1)$ is a homogeneous differential equation.
By Solution to Homogeneous Differential Equation, its solution is:
:$\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
where:
:$\map f {x, y} = \dfrac {3 x y + 2 y^2} {x^2}$
== Sources ==
* : $\S 2.7$: Homogeneous Equations: Problem $1 \ \text{(b)}$
Category:Examples of Homogeneous Differential Equation
Category:Examples of First Order ODEs
First Order ODE/x^2 y' = 3 (x^2 + y^2) arctan (y over x) + x y

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad x^2 \dfrac {\d y} {\d x} = 3 \paren {x^2 + y^2} \arctan \dfrac y x + x y$
is a homogeneous differential equation with solution:
:$y = x \tan C x^3$
== Proof ==
Let:
:$\map M {x, y} = 3 \paren {x^2 + y^2} \arctan \dfrac y x + x y$
:$\map N {x, y} = x^2$
Put $t x, t y$ for $x, y$:
Thus both $M$ and $N$ are homogeneous functions of degree $2$.
Thus, by definition, $(1)$ is a homogeneous differential equation.
By Solution to Homogeneous Differential Equation, its solution is:
:$\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
where:
:$\map f {x, y} = \dfrac {3 \paren {x^2 + y^2} \arctan \dfrac y x + x y} {x^2}$
Thus:
Substituting $u = \arctan z$:
Hence:
== Sources ==
* : $\S 2.7$: Homogeneous Equations: Problem $1 \ \text{(c)}$
Category:Examples of Homogeneous Differential Equation
Category:Examples of First Order ODEs
First Order ODE/x dy - y dx = (1 + y^2) dy

== Theorem ==
The first order ODE:
:$(1): \quad x \rd y - y \rd x = \paren {1 + y^2} \rd y$
has the general solution:
:$\dfrac x y = \dfrac 1 y - y + C$
== Proof ==
Rearranging, we have:
:$\dfrac {y \rd x - x \rd y} {y^2} = -\paren {\dfrac 1 {y^2} + 1} \rd y$
From the Quotient Rule for Derivatives:
:$\map \d {\dfrac x y} = \dfrac {y \rd x - x \rd y} {y^2}$
from which:
:$\map \d {\dfrac x y} = -\paren {\dfrac 1 {y^2} + 1} \rd y$
Hence the result:
:$\dfrac x y = \dfrac 1 y - y + C$
== Sources ==
* : $\S 2.9$: Integrating Factors: Problem $4 \ \text{(a)}$
Category:Examples of First Order ODEs
First Order ODE/x dy = (x^5 + x^3 y^2 + y) dx

== Theorem ==
The first order ODE:
:$(1): \quad x \rd y = \paren {x^5 + x^3 y^2 + y} \rd x$
has the general solution:
:$\arctan \dfrac x y = -\dfrac {x^4} 4 + C$
== Proof ==
Rearranging, we have:
:$y \rd x - x \rd y = -\paren {x^2 + y^2} x^3 \rd x$
from which:
:$\dfrac {y \rd x - x \rd y} {x^2 + y^2} = - x^3 \rd x$
From Differential of Arctangent of Quotient:
:$\map \d {\arctan \dfrac x y} = \dfrac {y \rd x - x \rd y} {x^2 + y^2}$
from which $(1)$ evolves into:
:$\map \d {\arctan \dfrac x y} = -x^3 \rd x$
Hence the result:
:$\arctan \dfrac x y = -\dfrac {x^4} 4 + C$
== Sources ==
* : $\S 2.9$: Integrating Factors: Problem $4 \ \text{(c)}$
Category:Examples of First Order ODEs
First Order ODE/x dy = (y + x^2 + 9 y^2) dx

== Theorem ==
The first order ODE:
:$(1): \quad x \rd y = \paren {y + x^2 + 9 y^2} \rd x$
has the general solution:
:$\map \arctan {\dfrac {3 y} x} = 3 x + C$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 2.9$: Integrating Factors: Problem $4 \ \text{(e)}$
Category:Examples of First Order ODEs
Category:First Order ODE/x dy = (y + x^2 + 9 y^2) dx
First Order ODE/x dy = (y + x^2 + 9 y^2) dx/Proof 1

== Theorem ==
== Proof ==
Divide both sides of $(1)$ by $x^2 \rd x$ to get:
:$\dfrac 1 x \dfrac {\d y} {\d x} = \dfrac 1 x \paren {\dfrac y x } + 1 + 9 \paren {\dfrac y x}^2$
Now apply the substitution:
:$y = u x$
This implies then that:
:$\dfrac {\d y} {\d x} = u + x \dfrac {\d u} {\d x}$
Now substitute everything into $(1)$ to get:
Now it becomes Solution to Separable Differential Equation and we end up with:
Substitute back for $u$:
:$\tan^{-1} \paren {\dfrac {3 y} x} = 3 x + C$
Category:First Order ODE/x dy = (y + x^2 + 9 y^2) dx
First Order ODE/x dy = (y + x^2 + 9 y^2) dx/Proof 2

== Theorem ==
== Proof ==
Let $z = \map \arctan {3y / x}$.
Then:
:$\dfrac {\partial z} {\partial x} = \dfrac 1 {1 + \paren {3 y / x}^2} \dfrac {-3 y} {x^2} = \dfrac {-3 y} {x^2 + 9 y^2}$
:$\dfrac {\partial z} {\partial y} = \dfrac 1 {1 + \paren {3 y / x}^2} \dfrac 3 x = \dfrac 3 {x^2 + 9 y^2}$
So:
:$\d z = \dfrac {3 x \rd y - 3 y \rd x} {x^2 + 9 y^2}$
Multiplying $(1)$ by $3$ and manipulating:
:$\dfrac {3 x \rd y - 3 y \rd x} {x^2 + 9 y^2} = 3 \rd x$
From Differential of Arctangent of Quotient:
:$\map \d {\map \arctan {\dfrac {3 y} x} } = 3 \rd x$
and so
:$\map \arctan {\dfrac {3 y} x} = 3 x + C$
Category:First Order ODE/x dy = (y + x^2 + 9 y^2) dx
Category:Integrating Factors
First Order ODE/x dy = (y + y^3) dx

== Theorem ==
The first order ODE:
:$(1): \quad x \dfrac {\d y} {\d x} = y + y^3$
has the general solution:
:$y = \dfrac x {\sqrt {C^2 - x^2} }$
== Proof ==
Category:Examples of First Order ODEs
Category:Examples of Solutions to Separable Differential Equation
First Order ODE/x dy = k y dx

== Theorem ==
The first order ODE:
:$(1): \quad x \rd y = k y \rd x$
has the general solution:
:$y = C x^k$
== Proof ==
$(1)$ can be expressed as:
Category:Examples of First Order ODEs
Category:Examples of Solutions to Separable Differential Equation
Category:First Order ODE/x dy = k y dx
First Order ODE/x sine (y over x) y' = y sine (y over x) + x

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad x \sin \dfrac y x \dfrac {\d y} {\d x} = y \sin \dfrac y x + x$
is a homogeneous differential equation with solution:
:$\cos \dfrac y x + \ln C x = 0$
== Proof ==
Let:
:$\map M {x, y} = y \sin \dfrac y x + x$
:$\map N {x, y} = x \sin \dfrac y x$
Put $t x, t y$ for $x, y$:
Thus both $M$ and $N$ are homogeneous functions of degree $1$.
Thus, by definition, $(1)$ is a homogeneous differential equation.
By Solution to Homogeneous Differential Equation, its solution is:
:$\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
where:
:$\map f {x, y} = \dfrac {y \sin \dfrac y x + x} {x \sin \dfrac y x}$
Thus:
== Sources ==
* : $\S 2.7$: Homogeneous Equations: Problem $1 \ \text{(d)}$
Category:Examples of Homogeneous Differential Equation
Category:Examples of First Order ODEs
First Order ODE/x y' = 2 y

== Theorem ==
The first order ODE:
:$x y' = 2 y$
has the general solution:
:$y = C x^2$
where $C$ is an arbitrary constant.
== Proof ==
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation: $(3.15)$
Category:First Order ODE/x dy = k y dx
First Order ODE/x y' = Root of (x^2 + y^2)

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad x y' = \sqrt {x^2 + y^2}$
is a homogeneous differential equation with solution:
:$3 x^2 \ln x = y \sqrt {x^2 + y^2} + x^2 \map \ln {y + \sqrt {x^2 + y^2} } + y^2 + C x^2$
== Proof ==
We divide through by $x$ to show that $(1)$ is homogeneous:
By Solution to Homogeneous Differential Equation, its solution is:
:$\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
where:
:$\map f {x, y} = \dfrac {\sqrt {x^2 + y^2} } x$
Thus:
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $4$
Category:Examples of Homogeneous Differential Equation
Category:Examples of First Order ODEs
First Order ODE/x y' = y + 2 x exp (- y over x)

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad x \dfrac {\d y} {\d x} = y + 2 x e^{-y/x}$
is a homogeneous differential equation with solution:
:$e^{y / x} = \ln x^2 + C$
== Proof ==
Let:
:$\map M {x, y} = y + 2 x e^{-y/x}$
:$\map N {x, y} = x$
Put $t x, t y$ for $x, y$:
Thus both $M$ and $N$ are homogeneous functions of degree $1$.
Thus, by definition, $(1)$ is a homogeneous differential equation.
By Solution to Homogeneous Differential Equation, its solution is:
:$\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
where:
:$\map f {x, y} = \dfrac {y + 2 x e^{-y / x} } x$
Thus:
== Sources ==
* : $\S 2.7$: Homogeneous Equations: Problem $1 \ \text{(e)}$
Category:Examples of Homogeneous Differential Equation
Category:Examples of First Order ODEs
First Order ODE/x y dy = x^2 dy + y^2 dx

== Theorem ==
The first order ODE:
:$(1): \quad x y \rd y = x^2 \rd y + y^2 \rd x$
has the general solution:
:$y = x \ln y + C x$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $9$
Category:Examples of First Order ODEs
Category:Examples of Bernoulli's Equation
Category:Examples of Homogeneous Differential Equation
Category:First Order ODE/x y dy = x^2 dy + y^2 dx
First Order ODE/x y dy = x^2 dy + y^2 dx/Proof 1

== Theorem ==
== Proof ==
Let $(1)$ be rearranged as:
:$(2): \quad \dfrac {\d x} {\d y} - \dfrac x y = -\dfrac {x^2} {y^2}$
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:
:$\map P y = -\dfrac 1 y$
:$\map Q y = -\dfrac 1 {y^2}$
:$n = 2$
and so is an example of Bernoulli's equation.
By Solution to Bernoulli's Equation it has the general solution:
:$(3): \quad \ds \frac {\map \mu y} {x^{n - 1} } = \paren {1 - n} \int \map Q y \map \mu y \rd y + C$
where:
:$\map \mu y = e^{\paren {1 - n} \int \map P y \rd y}$
Thus $\map \mu y$ is evaluated:
and so substituting into $(3)$:
Category:First Order ODE/x y dy = x^2 dy + y^2 dx
First Order ODE/x y dy = x^2 dy + y^2 dx/Proof 2

== Theorem ==
== Proof ==
Let $(1)$ be rearranged as:
:$(2): \quad y^2 \rd x = \paren {x y - x^2} \rd y$
Let:
:$\map M {x, y} = y^2$
:$\map N {x, y} = x y - x^2$
Put $t x, t y$ for $x, y$:
Thus both $M$ and $N$ are homogeneous functions of degree $2$.
Thus, by definition, $(1)$ is a homogeneous differential equation:
:$\dfrac {\d x} {\d y} = \dfrac {x^2 - x y} {y^2}$
By Solution to Homogeneous Differential Equation, its solution is:
:$\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
where:
:$\map f {y, x} = \dfrac {x y - x^2} {y^2}$
Hence:
Category:First Order ODE/x y dy = x^2 dy + y^2 dx
First Order ODE/y' + 2 x y = 1

== Theorem ==
The first order ODE:
:$y' + 2 x y = 1$
has the general solution:
:$y = e^{-{x^2} } \ds \int_a^x e^{t^2} \rd t$
where $a$ is an arbitrary constant.
== Proof ==
This is a linear first order ODE in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where:
:$\map p x = 2 x$
:$\map Q x = 1$
From Solution to Linear First Order Ordinary Differential Equation:
:$\ds y = e^{-\int P \rd x} \paren {\int Q e^{\int P \rd x} \rd x + C}$
Thus 
Further work on this is not trivial, as $\ds \int e^{x^2} \rd x$ has no solution in elementary functions.
== Sources ==
* : Chapter $\text I$: Existence of Solutions: $2$. Simple ideas about solutions: Example $1$.
Category:Examples of First Order ODEs
First Order ODE/y' + y = 0

== Theorem ==
The first order ODE:
:$\dfrac {\d y} {\d x} + y = 0$
has the general solution:
:$y = C e^{-x}$
where $C$ is an arbitrary constant.
== Proof ==
This first order ODE is in the form:
:$\dfrac {\d y} {\d x} + k y = 0$
where $k = 1$.
From First Order ODE: $\d y = k y \rd x$, this has the solution:
:$y = C e^{-x}$
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation: $(3.11)$
Category:First Order ODE/dy = k y dx
First Order ODE/y' - f (y) phi' (x) over f' (y) = phi (x) phi' (x) over f' (y)

== Theorem ==
Let $\map f y$ and $\map \phi x$ be known real functions of $y$ and $x$ respectively.
The general solution of:
:$(1): \quad \dfrac {\d y} {\d x} - \dfrac {\map f y} {\map {f'} y} \map {\phi'} x = \dfrac {\map \phi x \map {\phi'} x} {\map {f'} y}$
is:
:$\map f y = C e^{\map \phi x} - \map \phi x - 1$
== Proof ==
Let $u = \map f y$
Then by the Chain Rule for Derivatives:
:$\dfrac {\d u} {\d x} = \map {f'} y \dfrac {\d y} {\d x}$
Then:
This is a linear first order ordinary differential equation in the form:
:$\dfrac {\d u} {\d x} + \map P x u = \map Q x$
whose general solution from Solution to Linear First Order Ordinary Differential Equation is:
:$\ds u e^{\int P \rd x} = \int Q e^{\int P \rd x} \rd x + C$
In this instance:
:$\map P x = -\map {\phi'} x$
and:
:$\map Q x = \map \phi x \map {\phi'} x$
Thus:
Hence:
:$\ds u e^{-\map \phi x + A} = \int \map \phi x \map {\phi'} x e^{-\map \phi x + A} \rd x + C$
Let $v = \map \phi x$.
Then by Integration by Substitution:
This leaves us with:
:$u e^{-\map \phi x + A} = - e^A e^{-v} \paren {v + 1} + C$
subsuming $C'$ into $C$.
Substituting back in, the general solution is seen to be:
:$\map f y e^{-\map \phi x + A} = - e^{-\map \phi x + A} \paren {\map \phi x + 1} + C$
and so multiplying both sides by $e^{\map \phi x - A}$:
Hence the result.
== Sources ==
* : Exercise $18 \ (4)$
Category:Examples of First Order ODEs
First Order ODE/y' = (x + y)^2

== Theorem ==
The first order ODE:
:$\dfrac {\d y} {\d x} = \paren {x + y}^2$
has the general solution:
:$x + y = \map \tan {x + C}$
== Proof ==
Make the substitution:
:$z = x + y$
Then from First Order ODE in form $y' = f (a x + b y + c)$ with $a = b = 1$:
== Sources ==
* : $\S 2.7$: Homogeneous Equations: Problem $3 \ \text{(a)}$
Category:Examples of First Order ODEs
First Order ODE/y' = e^x

== Theorem ==
The first order ODE:
:$y' = e^x$
has the general solution:
:$y = e^x + C$
where $C$ is an arbitrary constant.
== Proof ==
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation: $(3.12)$
Category:First Order ODE/dy = k y dx