nyuuzyou/wikis huggingface dataset:
First Order ODE/(1 - x y) y' = y^2

== Theorem ==
The first order ODE:
:$(1): \quad \paren {1 - x y} y' = y^2$
has the general solution:
:$x y = \ln y + C$
== Proof ==
Let $(1)$ be rearranged as:
It can be seen that $(2)$ is a linear first order ODE in the form:
:$\dfrac {\d x} {\d y} + \map P y x = \map Q y$
where:
:$\map P y = \dfrac 1 y$
:$\map Q y = \dfrac 1 {y^2}$
Thus:
Thus from Solution by Integrating Factor, $(2)$ can be rewritten as:
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $2$
Category:Examples of Linear First Order ODEs
First Order ODE/(1 over x^3 y^2 + 1 over x) dx + (1 over x^2 y^3 - 1 over y) dy = 0

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad \paren {\dfrac 1 {x^3 y^2} + \dfrac 1 x} \rd x + \paren {\dfrac 1 {x^2 y^3} - \dfrac 1 y} \rd y = 0$
is an exact differential equation with solution:
:$-\dfrac 1 {2 x^2 y^2} = \ln \dfrac y x + C$
== Proof ==
Let:
:$\map M {x, y} = \dfrac 1 {x^3 y^2} + \dfrac 1 x$
:$\map N {x, y} = \dfrac 1 {x^2 y^3} - \dfrac 1 y$
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:
:$\map f {x, y} = -\dfrac 1 {2 x^2 y^2} + \ln x - \ln y$
and by Solution to Exact Differential Equation, the solution to $(1)$ is:
:$-\dfrac 1 {2 x^2 y^2} + \ln x - \ln y = C$
or:
:$-\dfrac 1 {2 x^2 y^2} = \ln \dfrac y x + C$
Category:Examples of Exact Differential Equation
Category:Examples of First Order ODEs
First Order ODE/(2 x + 3 y + 1) dx + (2 y - 3 x + 5) dy = 0

== Theorem ==
The first order ODE:
:$(1): \quad \paren {2 x + 3 y + 1} \rd x + \paren {2 y - 3 x + 5} \rd y = 0$
has the general solution:
:$3 \, \map \arctan {\dfrac {y + 1} {x - 1} } = \map \ln {\paren {y + 1}^2 + \paren {x - 1}^2} + C$
== Proof ==
Rewriting $(1)$ as:
:$\dfrac {\d y} {\d x} = -\dfrac {2 x + 3 y + 1} {-3 x + 2 y + 5}$
we note that it is in the form:
:$\dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$
where:
:$a e = -4 \ne b d = 9$
Hence we can use:
:First Order ODE in form $y' = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$
which can be solved by substituting:
:$x := z - h$
:$y := w - k$
where:
:$h = \dfrac {c e - b f} {a e - b d}$
:$k = \dfrac {a f - c d} {a e - b d}$
So let:
:$x = z - h$ where $h = \dfrac {\paren {1 \times 2} - \paren {3 \times 5} } {\paren {2 \times 2} - \paren {3 \times -3} } = \dfrac {-13} {13} = -1$
:$y = w - k$ where $k = \dfrac {\paren {2 \times 5} - \paren {1 \times -3} } {\paren {2 \times 2} - \paren {3 \times -3} } = \dfrac {13} {13} = 1$
Then $(1)$ is transformed into:
:$(2): \quad \dfrac {\d w} {\d z} = -\dfrac {2 z + 3 w} {3 z - 2 w}$
Let:
:$\map M {w, z} = 2 z + 3 w$
:$\map N {w, z} = 3 z - 2 w$
We have that:
:$\map M {t w, t z} = 2 t z + 3 t w = t \paren {2 z + 3 w} = t \, \map M {x, y}$
:$\map N {t w, t z} = 3 t z - 2 t w = t \paren {3 z - 2 w} = t \, \map N {x, y}$
Thus both $M$ and $N$ are homogeneous functions of degree $1$.
Thus by definition $(2)$ is a homogeneous differential equation.
So:
By Solution to Homogeneous Differential Equation:
:$\ds \ln z = \int \frac {\d u} {\map f {1, u} - u} + C$
where:
:$\map f {1, u} = \dfrac {2 + 3 u} {3 - 2 u}$
Hence:
Replacing all the substitutions:
:$\map \ln {x - 1} = \dfrac 3 2 \, \map \arctan {\dfrac {y + 1} {x - 1} } - \dfrac 1 2 \, \map \ln {1 + \paren {\dfrac {y + 1} {x - 1} } } + C$
Tidying up and reassigning constants appropriately:
:$3 \, \map \arctan {\dfrac {y + 1} {x - 1} } = \map \ln {\paren {y + 1}^2 + \paren {x - 1}^2} + C$
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $3$
Category:Examples of First Order ODEs
First Order ODE/(2 x y^3 + y cosine x) dx + (3 x^2 y^2 + sine x) dy

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad \paren {2 x y^3 + y \cos x} \d x + \paren {3 x^2 y^2 + \sin x} \d y = 0$
is an exact differential equation with solution:
:$x^2 y^3 + y \sin x = C$
This can also be presented as:
:$\dfrac {\d y} {\d x} = -\dfrac {2 x y^3 + y \cos x} {3 x^2 y^2 + \sin x}$
== Proof ==
Let:
:$\map M {x, y} = 2 x y^3 + y \cos x$
:$\map N {x, y} = 3 x^2 y^2 + \sin x$
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:
:$\map f {x, y} = x^2 y^3 + y \sin x$
and by Solution to Exact Differential Equation, the solution to $(1)$ is:
:$x^2 y^3 + y \sin x = C$
== Sources ==
* : $\S 2.8$: Exact Equations: Problem $10$
Category:Examples of Exact Differential Equation
Category:Examples of First Order ODEs
First Order ODE/(3 x^2 - y^2) dy - 2 x y dx = 0

== Theorem ==
The first order ODE:
:$(1): \quad \paren {3 x^2 - y^2} \rd y - 2 x y \rd x = 0$
has the general solution:
:$\dfrac 1 y - \dfrac {x^2} {y^3} = C$
This can also be presented in the form:
:$\dfrac {\d y} {\d x} = \dfrac {2 x y} {3 x^2 - y^2}$
== Proof ==
We note that $(1)$ is in the form:
:$\map M {x, y} \rd x + \map N {x, y} \rd y = 0$
but that $(1)$ is not exact.
So, let:
:$\map M {x, y} = -2 x y$
:$\map N {x, y} = 3 x^2 - y^2$
Let:
:$\map P {x, y} = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$
Thus:
It can be observed that:
Thus $\dfrac {\map P {x, y}} {\map M {x, y} }$ is a function of $y$ only.
So Integrating Factor for First Order ODE: Function of One Variable can be used:
:$\map \mu y = e^{\int -\map h y \rd y}$
Hence:
Thus an integrating factor for $(1)$ has been found:
:$\mu = \dfrac 1 {y^4}$
which yields, when multiplying it throughout $(1)$:
:$\paren {\dfrac {3 x^2} {y^4} - \dfrac 1 {y^2} } \rd y - \dfrac {2 x} {y^3} \rd x = 0$
which is now exact.
By First Order ODE: $\paren {\dfrac {3 x^2} {y^4} - \dfrac 1 {y^2} } \rd y - \dfrac {2 x} {y^3} \rd x = 0$, its solution is:
:$\dfrac 1 y - \dfrac {x^2} {y^3} = C$
== Sources ==
* : $\S 2.9$: Integrating Factors: Problem $2 \ \text{(a)}$
Category:Examples of First Order ODEs
First Order ODE/(3 x^2 over y^4 - 1 over y^2) dy - 2 x over y^3 dx = 0

== Theorem ==
The first order ODE:
:$(1): \quad \paren {\dfrac {3 x^2} {y^4} - \dfrac 1 {y^2} } \rd y - \dfrac {2 x} {y^3} \rd x = 0$
is an exact differential equation with general solution:
:$\dfrac 1 y - \dfrac {x^2} {y^3} = C$
== Proof ==
Let $M$ and $N$ be defined as:
:$\map M {x, y} = -\dfrac {2 x} {y^3}$
:$\map N {x, y} = \dfrac {3 x^2} {y^4} - \dfrac 1 {y^2}$
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:
and:
Thus:
:$\map f {x, y} = \dfrac 1 y - \dfrac {x^2} {y^3}$
and by Solution to Exact Differential Equation, the general solution to $(1)$ is:
:$\dfrac 1 y - \dfrac {x^2} {y^3} = C$
Category:Examples of Exact Differential Equation
Category:Examples of First Order ODEs
First Order ODE/(6x + 4y + 3) dx + (3x + 2y + 2) dy = 0

== Theorem ==
The first order ODE:
:$(1): \quad \paren {6 x + 4 y + 3} \rd x + \paren {3 x + 2 y + 2} \rd y = 0$
has the general solution:
:$3 x + 2 y + \map \ln {\paren {3 x + 2 y}^2} + x = C$
== Proof ==
We put $(1)$ in the form:
:$\dfrac {\d y} {\d x} = -\dfrac {6 x + 4 y + 3} {3 x + 2 y + 2}$
Substitute $z = 3 x + 2 y$, which gives:
:$\dfrac {\d z} {\d x} = 3 + 2 \dfrac {\d y} {\d x}$
and so:
:$\dfrac {\d z} {\d x} = -2 \dfrac {2 z + 3} {z + 2} + 3$
which simplifies down to:
:$\dfrac {\d z} {\d x} = \dfrac {-z} {z + 2}$
This is separable:
:$\ds -\int \frac {z + 2} z \rd z = \int \rd x$
which gives:
:$- z - 2 \ln z = x + C$
which, after the substitutions, gives:
:$3 x + 2 y + \map \ln {\paren {3 x + 2 y}^2} + x = C$
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $14$
Category:Examples of First Order ODEs
First Order ODE/(exp x - 3 x^2 y^2) y' + y exp x = 2 x y^3

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad \paren {e^x - 3 x^2 y^2} y' + y e^x = 2 x y^3$
is an exact differential equation with solution:
:$y e^x - x^2 y^3 = C$
== Proof ==
Let $(1)$ be expressed as:
:$\paren {y e^x - 2 x y^3} \rd x + \paren {e^x - 3 x^2 y^2} \rd y = 0$
Let:
:$\map M {x, y} = y e^x - 2 x y^3$
:$\map N {x, y} = e^x - 3 x^2 y^2$
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:
and:
Thus:
:$\map f {x, y} = y e^x + x^2 y^3$
and by Solution to Exact Differential Equation, the solution to $(1)$ is:
:$y e^x + x^2 y^3 = C$
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $10$
Category:Examples of Exact Differential Equation
Category:Examples of First Order ODEs
First Order ODE/(exp y - 2 x y) y' = y^2

== Theorem ==
The first order ODE:
:$(1): \quad \paren {e^y - 2 x y} y' = y^2$
has the general solution:
:$x y^2 = e^y + C$
== Proof ==
Let $(1)$ be rearranged as:
:$\dfrac {\d y} {\d x} = \dfrac {y^2} {e^y - 2 x y}$
Hence:
:$(2): \quad \dfrac {\d x} {\d y} + \dfrac 2 y x = \dfrac {e^y} {y^2}$
It can be seen that $(2)$ is a linear first order ODE in the form:
:$\dfrac {\d x} {\d y} + \map P y x = \map Q y$
where:
:$\map P y = \dfrac 2 y$
:$\map Q y = \dfrac {e^y} {y^2}$
Thus:
Thus from Solution by Integrating Factor, $(2)$ can be rewritten as:
:$\map {\dfrac \d {\d y} } {x y^2} = e^y$
and the general solution is:
:$x y^2 = e^y + C$
== Sources ==
* : $\S 2.10$: Problem $4 \ \text{(a)}$
Category:Examples of First Order ODEs
First Order ODE/(sine x sine y - x e^y) dy = (e^y + cosine x cosine y) dx

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad \paren {\sin x \sin y - x e^y} \rd y = \paren {e^y + \cos x \cos y} \rd x$
is an exact differential equation with solution:
:$\sin x \cos y + x e^y = C$
This can also be presented as:
:$\dfrac {\d y} {\d x} = \dfrac {e^y + \cos x \cos y} {\sin x \sin y - x e^y}$
== Proof ==
First express $(1)$ in the form:
:$(2): \quad -\paren {e^y + \cos x \cos y} + \paren {\sin x \sin y - x e^y} \dfrac {\d y} {\d x}$
Let:
:$\map M {x, y} = -\paren {e^y + \cos x \cos y}$
:$\map N {x, y} = \sin x \sin y - x e^y$
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:
:$\map f {x, y} = -\sin x \cos y - x e^y$
and by Solution to Exact Differential Equation, the solution to $(1)$, after simplification, is:
:$\sin x \cos y + x e^y = C$
== Sources ==
* : $\S 2.8$: Exact Equations: Problem $7$
Category:Examples of Exact Differential Equation
Category:Examples of First Order ODEs
First Order ODE/(x^2 - 2 y^2) dx + x y dy = 0

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad \paren {x^2 - 2 y^2} \rd x + x y \rd y = 0$
is a homogeneous differential equation with solution:
:$y^2 = x^2 + C x^4$
== Proof ==
$(1)$ can also be rendered:
:$\dfrac {\d y} {\d x} = -\dfrac {x^2 - 2 y^2} {x y}$
Let:
:$\map M {x, y} = x^2 - 2 y^2$
:$\map N {x, y} = x y$
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:
:$\map f {x, y} = -\dfrac {x^2 - 2 y^2} {x y}$
Hence:
== Sources ==
* : $\S 2.7$: Homogeneous Equations: Problem $1 \ \text{(a)}$
Category:Examples of Homogeneous Differential Equation
Category:Examples of First Order ODEs
First Order ODE/(x^2 y^3 + y) dx = (x^3 y^2 - x) dy

== Theorem ==
The first order ODE:
:$(1): \quad \paren {x^2 y^3 + y} \rd x = \paren {x^3 y^2 - x} \rd y$
has the general solution:
:$-\dfrac 1 {2 x^2 y^2} = \ln \dfrac y x + C$
== Proof ==
Let $(1)$ be expressed as:
:$(2): \quad \paren {y + x^2 y^3} \rd x + \paren {x - x^3 y^2} \rd y = 0$
We note that $(2)$ is in the form:
:$\map M {x, y} \rd x + \map N {x, y} \rd y = 0$
but is not exact.
So, let:
:$\map M {x, y} = y + x^2 y^3$
:$\map N {x, y} = x - x^3 y^2$
Let:
:$\map P {x, y} = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$
Thus:
It can be observed that:
Thus $\dfrac {\map P {x, y} } {\map M {x, y} }$ is a function of $x y$.
So Integrating Factor for First Order ODE: Function of Product of Variables can be used.
Let $z = x y$.
Then:
:$\map \mu {x y} = \map \mu z = e^{\int \map g z \rd z}$
Hence:
Thus an integrating factor for $(1)$ has been found:
:$\mu = \dfrac 1 {x^3 y^3}$
which yields, when multiplying it throughout $(2)$:
:$\paren {\dfrac 1 {x^3 y^2} + \dfrac 1 x} \rd x + \paren {\dfrac 1 {x^2 y^3} - \dfrac 1 y} \rd y = 0$
which is now exact.
By First Order ODE: $\paren {\dfrac 1 {x^3 y^2} + \dfrac 1 x} \rd x + \paren {\dfrac 1 {x^2 y^3} - \dfrac 1 y} \rd y = 0$, its general solution is:
:$-\dfrac 1 {2 x^2 y^2} = \ln \dfrac y x + C$
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $6$
Category:Examples of First Order ODEs
First Order ODE/(x + (2 over y)) dy + y dx = 0

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad \paren {x + \dfrac 2 y} \rd y + y \rd x = 0$
is an exact differential equation with solution:
:$x y + 2 \ln y = C$
This can also be presented as:
:$\dfrac {\d y} {\d x} + \dfrac y {x + \dfrac 2 y} = 0$
== Proof ==
Let:
:$\map M {x, y} = y$
:$\map N {x, y} = x + \dfrac 2 y$
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:
:$\map f {x, y} = x y + 2 \ln y$
and by Solution to Exact Differential Equation, the solution to $(1)$ is:
:$x y + 2 \ln y = C$
== Sources ==
* : $\S 2.8$: Exact Equations: Problem $1$
Category:Examples of Exact Differential Equation
Category:Examples of First Order ODEs
First Order ODE/(x + y) dx = (x - y) dy

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad \paren {x + y} \rd x = \paren {x - y} \rd y$
is a homogeneous differential equation with general solution:
:$\arctan \dfrac y x = \ln \sqrt {x^2 + y^2} + C$
== Proof 1 ==
== Proof 2 ==
Category:Examples of First Order ODEs
Category:Examples of Homogeneous Differential Equation
Category:First Order ODE/(x + y) dx = (x - y) dy
First Order ODE/(x + y) dx = (x - y) dy/Proof 1

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad \paren {x + y} \rd x = \paren {x - y} \rd y$
is a homogeneous differential equation with solution:
:$\arctan \dfrac y x = \ln \sqrt{x^2 + y^2} + C$
== Proof ==
Let:
:$\map M {x, y} = x + y$
:$\map N {x, y} = x - y$
We have that:
:$\map M {t x, t y} = t x + t y = t \paren {x + y} = t \map M {x, y}$
:$\map N {t x, t y} = t x - t y = t \paren {x - y} = t \map N {x, y}$
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:
:$\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
where:
:$\map f {1, z} = \dfrac {1 + z} {1 - z}$
Hence:
Substituting $y / x$ for $z$ reveals the solution:
:$\arctan \dfrac y x = \ln \sqrt{x^2 + y^2} + C$
== Sources ==
* : $\S 2.7$: Homogeneous Equations: Example $1$
Category:Examples of First Order ODEs
First Order ODE/(x + y + 4) over (x + y - 6)

== Theorem ==
The first order ODE:
:$(1): \quad \dfrac {\d y} {\d x} = \dfrac {x + y + 4} {x + y - 6}$
has the general solution:
:$y - x = 5 \, \map \ln {x + y - 1} + C$
== Proof ==
We note that $(1)$ is in the form:
:$\dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$
where:
:$a e = b d = 1$
Hence we can use First Order ODE in form $y' = \map F {\dfrac{a x + b y + c} {d x + e y + f} }$ where $a e = b d$.
Let:
:$z = x + y$
to obtain:
:$\dfrac {\d z} {\d x} = b \, \map F {\dfrac {a z + a c} {d z + f} } + a$
where:
:$a = 1$
:$c = 4$
:$d = 1$
:$f = -6$
which gives:
Substitute $v = z - 1$ which gives:
:$\dfrac {\d z} {\d v} = 1$
and thence:
:$v = x + y - 1$:
Substituting $x + y - 1$ for $v$:
== Sources ==
* : $\S 2.7$: Homogeneous Equations: Problem $5 \ \text{(a)}$
Category:Examples of First Order ODEs
First Order ODE/(x + y + 4) over (x - y - 6)

== Theorem ==
The first order ODE:
:$(1): \quad \dfrac {\d y} {\d x} = \dfrac {x + y + 4} {x - y - 6}$
has the general solution:
:$\map \arctan {\dfrac {y + 5} {x - 1} } = \ln \sqrt {\paren {x - 1}^2 + \paren {y + 5}^2} + C$
== Proof ==
We note that $(1)$ is in the form:
:$\dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$
where:
:$a e = -1 \ne b d = 1$
Hence we can use First Order ODE in form $y' = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$.
Let:
:$x = z - h$ where $h = 2 / -2 = -1$
:$y = w - k$ where $k = -10 / -2 = 5$.
Then:
:$\dfrac {\d w} {\d z} = \dfrac {z + w} {z - w}$
From the solution to $\dfrac {\d y} {\d x} = \dfrac {x + y} {x - y}$:
:$\arctan \dfrac w z = \ln \sqrt {z^2 + w^2} + C$
We have:
:$x = z - \paren {-1}$
:$y = w - 5$
and so:
:$z = x - 1$
:$w = y + 5$
This gives:
:$\map \arctan {\dfrac {y + 5} {x - 1} } = \ln \sqrt {\paren {x - 1}^2 + \paren {y + 5}^2} + C$
== Sources ==
* : $\S 2.7$: Homogeneous Equations: Problem $5 \ \text{(a)}$
Category:Examples of First Order ODEs
First Order ODE/(x exp y + y - x^2) dy = (2 x y - exp y - x) dx

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad \paren {x e^y + y - x^2} \rd y = \paren {2 x y - e^y - x} \rd x$
is an exact differential equation with solution:
:$2 x e^y + x^2 + y^2 - 2 x^2 y = C$
== Proof ==
Let $(1)$ be expressed in the form:
:$\paren {e^y + x - 2 x y} \rd x + \paren {x e^y + y - x^2} rd y = 0$
Let:
:$\map M {x, y} = e^y + x - 2 x y$
:$\map N {x, y} = x e^y + y - x^2$
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:
and:
Thus:
:$\map f {x, y} = x e^y + \dfrac {x^2 + y^2} 2 - x^2 y$
and by Solution to Exact Differential Equation, the solution to $(1)$ is:
:$2 x e^y + x^2 + y^2 - 2 x^2 y = C$
after multiplying by $2$ to clear the fraction.
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $24$
Category:Examples of Exact Differential Equation
Category:Examples of First Order ODEs
First Order ODE/(x y - 1) dx + (x^2 - x y) dy = 0

== Theorem ==
The first order ODE:
:$(1): \quad \paren {x y - 1} \rd x + \paren {x^2 - x y} \rd y = 0$
has the general solution:
:$x y - \ln x - \dfrac {y^2} 2 + C$
This can also be presented in the form:
:$\dfrac {\d y} {\d x} + \dfrac {x y - 1} {x^2 - x y} = 0$
== Proof ==
We note that $(1)$ is in the form:
:$\map M {x, y} \rd x + \map N {x, y} \rd y = 0$
but that $(1)$ is not exact.
So, let:
:$\map M {x, y} = x y - 1$
:$\map N {x, y} = x^2 - x y = x \paren {x - y}$
Let:
:$\map P {x, y} = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$
Thus:
It can be observed that:
Thus $\dfrac {\map P {x, y} } {\map N {x, y}}$ is a function of $x$ only.
So Integrating Factor for First Order ODE: Function of One Variable can be used:
:$\map \mu y = e^{\int \map g x \rd x}$
Hence:
Thus an integrating factor for $(1)$ has been found:
:$\mu = \dfrac 1 x$
which yields, when multiplying it throughout $(1)$:
:$\paren {y - \dfrac 1 x} \rd x + \paren {x - y} \rd y = 0$
which is now exact.
By First Order ODE: $\paren {y - \dfrac 1 x} \rd x + \paren {x - y} \rd y = 0$, its solution is:
:$x y - \ln x - \dfrac {y^2} 2 + C$
== Sources ==
* : $\S 2.9$: Integrating Factors: Problem $2 \ \text{(b)}$
Category:Examples of First Order ODEs
First Order ODE/(y^2 - 3 x y - 2 x^2) dx = (x^2 - x y) dy

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad \paren {y^2 - 3 x y - 2 x^2} \rd x = \paren {x^2 - x y} \rd y$
is a homogeneous differential equation with solution:
:$y^2 x^2 - 2 y x^3 + x^4 = C$
== Proof ==
$(1)$ can also be rendered:
:$\dfrac {\rd y} {\rd x} = \dfrac {y^2 - 3 x y - 2 x^2} {x^2 - x y}$
Let:
:$\map M {x, y} = y^2 - 3 x y - 2 x^2$
:$\map N {x, y} = x^2 - x y$
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:
:$\map f {x, y} = \dfrac {y^2 - 3 x y - 2 x^2} {x^2 - x y}$
Hence:
Substituting back for $z$ and tidying up, the result is obtained:
:$y^2 x^2 - 2 y x^3 + x^4 = C$
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $20$
Category:Examples of Homogeneous Differential Equation
Category:Examples of First Order ODEs
First Order ODE/(y^2 exp x y + cosine x) dx + (exp x y + x y exp x y) dy = 0

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad \paren {y^2 e^{x y} + \cos x} \rd x + \paren {e^{x y} + x y e^{x y} } \rd y = 0$
is an exact differential equation with solution:
:$y e^{x y} + \sin x = C$
== Proof ==
Let:
:$\map M {x, y} = y^2 e^{x y} + \cos x$
:$\map N {x, y} = e^{x y} + x y e^{x y}$
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:
:$\map f {x, y} = y e^{x y} + \sin x$
and by Solution to Exact Differential Equation, the solution to $(1)$ is:
:$y e^{x y} + \sin x = C$
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $17$
Category:Examples of Exact Differential Equation
Category:Examples of First Order ODEs
First Order ODE/(y + y cosine x y) dx + (x + x cosine x y) dy = 0

== Theorem ==
The first order ordinary differential equation:
:$(1): \quad \paren {y + y \cos x y} \rd x + \paren {x + x \cos x y} \rd y = 0$
is an exact differential equation with solution:
:$x y + \sin x y = C$
This can also be presented as:
:$\dfrac {\d y} {\d x} + \dfrac {y + y \cos x y} {x + x \cos x y} = 0$
== Proof ==
Let:
:$\map M {x, y} = y + y \cos x y$
:$\map N {x, y} = x + x \cos x y$
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:
:$\map f {x, y} = x y + \sin x y$
and by Solution to Exact Differential Equation, the solution to $(1)$ is:
:$x y + \sin x y = C$
== Sources ==
* : $\S 2.8$: Exact Equations: Problem $5$
Category:Examples of Exact Differential Equation
Category:Examples of First Order ODEs
First Order ODE/(y - 1 over x) dx + (x - y) dy = 0

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