avewright/proofwiki-math huggingface dataset:
Theorem: First Order ODE/(x + y) dx = (x - y) dy

is a homogeneous differential equation with general 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.
{{begin-eqn}}
{{eqn | l = \frac {\d y} {\d x}
      | r = \frac {x + y} {x - y}
      | c = 
}}
{{eqn | r = \frac {1 + y / x} {1 - y / x}
      | c = dividing top and bottom by $x$
}}
{{eqn | r = \frac {1 + z} {1 - z}
      | c = substituting $z$ for $y / x$
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \ln x
      | r = \int \frac {\d z} {\dfrac {1 + z} {1 - z} - z}
      | c = 
}}
{{eqn | r = \int \frac {\paren {1 - z} \rd z} {1 + z - z \paren {1 - z} }
      | c = 
}}
{{eqn | r = \int \frac {\paren {1 - z} \rd z} {1 + z^2}
      | c = 
}}
{{eqn | r = \int \frac {\d z} {1 + z^2} - \int \frac {z \rd z} {1 + z^2}
      | c = 
}}
{{eqn | r = \arctan z - \int \frac {z \rd z} {1 + z^2} + C
      | c = Primitive of $\dfrac 1 {x^2 + a^2}$
}}
{{eqn | r = \arctan z - \frac 1 2 \map \ln {1 + z^2} + C
      | c = Primitive of $\dfrac x {x^2 + a^2}$
}}
{{end-eqn}}
Substituting $y / x$ for $z$ reveals the solution:
:$\arctan \dfrac y x = \ln \sqrt{x^2 + y^2} + C$
{{qed}}
Theorem: First Order ODE/(x + y) dx = (x - y) dy

is a homogeneous differential equation with general solution:
:$\arctan \dfrac y x = \ln \sqrt {x^2 + y^2} + C$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \paren {x + y} \rd x
      | r = \paren {x - y} \rd y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x \rd y - y \rd x
      | r = x \rd x + y \rd y
      | c = rearranging
}}
{{eqn | ll= \leadsto
      | l = \frac {x \rd y - y \rd x} {x^2 + y^2}
      | r = \frac {x \rd x + y \rd y} {x^2 + y^2}
      | c = dividing through by $x^2 + y^2$
}}
{{eqn | ll= \leadsto
      | l = \frac {x \rd y - y \rd x} {x^2 + y^2}
      | r = \frac {\map \d {x^2 + y^2} } {2 \paren {x^2 + y^2} }
      | c = Differential of Sum of Squares
}}
{{eqn | ll= \leadsto
      | l = \map \d {\arctan \dfrac y x}
      | r = \frac {\map \d {x^2 + y^2} } {2 \paren {x^2 + y^2} }
      | c = Differential of Arctangent of Quotient
}}
{{eqn | ll= \leadsto
      | l = \arctan \dfrac y x
      | r = \frac {\map \ln {x^2 + y^2} } 2 + C
      | c = integrating
}}
{{eqn | ll= \leadsto
      | l = \arctan \dfrac y x
      | r = \ln \sqrt {x^2 + y^2} + C
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: First Order ODE/(x^2 - 2 y^2) dx + x y dy = 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 $x, y$:
{{begin-eqn}}
{{eqn | l = \map M {t x, t y}
      | r = \paren {t x}^2 - 2 \paren {t y}^2
      | c = 
}}
{{eqn | r = t^2 \paren {x^2 - 2 y^2}
      | c = 
}}
{{eqn | r = t^2 \, \map M {x, y}
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map N {t x, t y}
      | r = t x t y
      | c = 
}}
{{eqn | r = t^2 \paren {x y}
      | c = 
}}
{{eqn | r = t^2 \, \map N {x, y}
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \ln x
      | r = \int \frac {\d z} {-\frac {1 - 2 z^2} {z} - z} + C_1
      | c = 
}}
{{eqn | r = \int \frac {z \rd z} {z^2 - 1} + C_1
      | c = 
}}
{{eqn | r = \frac 1 2 \, \map \ln {z^2 - 1} + C_1
      | c = Primitive of $\dfrac x {x^2 - a^2}$
}}
{{eqn | ll= \leadsto
      | l = \map \ln {z^2 - 1}
      | r = \map \ln {x^2} + \ln C
      | c = 
}}
{{eqn | ll= \leadsto
      | l = z^2 - 1
      | r = C x^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {y^2} {x^2} - 1
      | r = C x^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y^2
      | r = x^2 + C x^4
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: First Order ODE/x^2 y' - 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 $x, y$:
{{begin-eqn}}
{{eqn | l = \map M {t x, t y}
      | r = 3 t x t y + 2 \paren {t y}^2
      | c = 
}}
{{eqn | r = t^2 \paren {3 x y + 2 y^2}
      | c = 
}}
{{eqn | r = t^2 \, \map M {x, y}
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map N {t x, t y}
      | r = \paren {t x}^2
      | c = 
}}
{{eqn | r = t^2 \paren {x^2}
      | c = 
}}
{{eqn | r = t^2 \, \map N {x, y}
      | c = 
}}
{{end-eqn}}
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:
:$\map f {x, y} = \dfrac {3 x y + 2 y^2} {x^2}$
{{begin-eqn}}
{{eqn | l = \ln x
      | r = \int \frac {\d z} {\dfrac {3 z + 2 z^2} {1^2} - z} + C_1
      | c = 
}}
{{eqn | r = \int \frac {\d z} {2 z \paren {1 + z} } + C_1
      | c = 
}}
{{eqn | r = \frac 1 2 \ln \frac z {z + 1} + \ln C_2
      | c = Primitive of $\dfrac 1 {x \paren {a x + b} }$
}}
{{eqn | ll= \leadsto
      | l = \ln x^2
      | r = \ln \frac z {z + 1} + \ln C_3
      | c = 
}}
{{eqn | r = \ln \frac {C_3 z} {z + 1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x^2
      | r = \frac {C_3 y / x} {y / x + 1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = C x^2 \paren {x + y}
      | c = where $C := \dfrac 1 {C_3}$
}}
{{end-eqn}}
{{qed}}
Theorem: Set of Condensation Points is Monotone

Let $T = \struct {S, \tau}$ be a topological space.
Let $A, B$ be subsets of $S$.
Then:
:$A \subseteq B \implies {A^0} \subseteq B^0$
where
:$A^0$ denotes the set of condensation points of $A$

Proof:
Assume
:$A \subseteq B$
Let $x \in A^0$.
By definition of set of condensation points:
:$x$ is condensation point of $A$
By definition of condensation point:
:$x$ is limit point of $A$ such that $\forall U \in \tau: A \cap U$ is uncountable
Thus by Limit Point of Subset is Limit Point of Set:
:$x$ is limit point of $B$
Let $U \in \tau$.
By definition of condensation point:
:$A \cap U$ is uncountable
By Set Intersection Preserves Subsets/Corollary:
:$A \cap U \subseteq B \cap U$
Thus by Subset of Countable Set is Countable
:$B \cap U$ is uncountable
Then by definition:
:$x$ is condensation point of $B$
Thus by definition of set of condensation points:
:$x \in B^0$
{{qed}}
Theorem: First Order ODE/x^2 y' = 3 (x^2 + y^2) arctan (y over 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 $x, y$:
{{begin-eqn}}
{{eqn | l = \map M {t x, t y}
      | r = 3 \paren {\paren {t x}^2 + \paren {t y}^2} \arctan \dfrac {t y} {t x} + t x t y
      | c = 
}}
{{eqn | r = t^2 \paren {3 \paren {x^2 + y^2} \arctan \dfrac y x + x y}
      | c = 
}}
{{eqn | r = t^2 \, \map M {x, y}
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map N {t x, t y}
      | r = \paren {t x}^2
      | c = 
}}
{{eqn | r = t^2 \paren {x^2}
      | c = 
}}
{{eqn | r = t^2 \, \map N {x, y}
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \ln x
      | r = \int \frac {\d z} {\dfrac {3 \paren {1^2 + z^2} \arctan \frac z 1 + z} {1^2} - z} + C_1
      | c = 
}}
{{eqn | r = \int \frac {\d z} {3 \paren {1 + z^2} \arctan z} + C_1
      | c = 
}}
{{end-eqn}}
Substituting $u = \arctan z$:
{{begin-eqn}}
{{eqn | l = \dfrac {\d u} {\d x}
      | r = \frac 1 {z^2 + 1}
      | c = Derivative of Arctangent Function
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\d z} {\d u}
      | r = z^2 + 1
      | c = 
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \ln x
      | r = \int \frac {\d z} {3 \paren {1 + z^2} \arctan z} + C_1
      | c = 
}}
{{eqn | r = \int \frac {\d u} {3 u} + C_1
      | c = 
}}
{{eqn | r = \frac 1 3 \ln u + C_1
      | c = Primitive of Reciprocal
}}
{{eqn | ll= \leadsto
      | l = \ln x^3 + \ln C
      | r = \ln u
      | c = putting $\ln C = - C_1$
}}
{{eqn | ll= \leadsto
      | l = C x^3
      | r = u
      | c = 
}}
{{eqn | ll= \leadsto
      | l = C x^3
      | r = \arctan \frac y x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = x \tan C x^3
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: First Order ODE/x sine (y over x) y' = y sine (y over 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 $x, y$:
{{begin-eqn}}
{{eqn | l = \map M {t x, t y}
      | r = t y \sin \dfrac t y t x + t x
      | c = 
}}
{{eqn | r = t \paren {y \sin \dfrac y x + x}
      | c = 
}}
{{eqn | r = t \, \map M {x, y}
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map N {t x, t y}
      | r = t x \sin \dfrac {t y} {t x}
      | c = 
}}
{{eqn | r = t \paren {x \sin \dfrac y x}
      | c = 
}}
{{eqn | r = t \, \map N {x, y}
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \ln x
      | r = \int \frac {\d z} {\dfrac {z \sin z + 1} {\sin z} - z} + C_1
      | c = 
}}
{{eqn | r = \int \sin z \rd z + C_1
      | c = 
}}
{{eqn | r = -\cos z + C_1
      | c = Primitive of $\sin z$
}}
{{eqn | ll= \leadsto
      | l = \ln x + \ln C
      | r = -\cos z
      | c = putting $\ln C = - C_1$
}}
{{eqn | ll= \leadsto
      | l = \cos \frac y x + \ln C x = 0
      | r = 0
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: First Order ODE/x y' = y + 2 x exp (- y over 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 $x, y$:
{{begin-eqn}}
{{eqn | l = \map M {t x, t y}
      | r = t y + 2 t x e^{-t y / t x}
      | c = 
}}
{{eqn | r = t \paren {y + 2 x e^{-y / x} }
      | c = 
}}
{{eqn | r = t \, \map M {x, y}
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map N {t x, t y}
      | r = t x
      | c = 
}}
{{eqn | r = t \, \map N {x, y}
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \ln x
      | r = \int \frac {\d z} {\dfrac {z + 2 e^{-z} } 1 - z} + C_1
      | c = 
}}
{{eqn | r = \int \frac {e^z} 2 \rd z + C_1
      | c = 
}}
{{eqn | r = \frac {e^z} 2 + C_1
      | c = Primitive of Exponential Function
}}
{{eqn | ll= \leadsto
      | l = 2 \ln x + C
      | r = e^z
      | c = putting $C = -\dfrac {C_1} 2$
}}
{{eqn | ll= \leadsto
      | l = e^{y / x}
      | r = \ln x^2 + C
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: First Order ODE in form y' = f (a x + b y + c)

The first order ODE:
:$\dfrac {\d y} {\d x} = \map f {a x + b y + c}$
can be solved by substituting:
:$z := a x + b y + c$
to obtain:
:$\ds x = \int \frac {\d z} {b \map f z + a}$

Proof:
We have:
:$\dfrac {\d y} {\d x} = \map f {a x + b y + c}$
Put:
: $z := a x + b y + c$
Then:
{{begin-eqn}}
{{eqn | l = z
      | r = a x + b y + c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = a + b \dfrac {\d y} {\d x}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\d y} {\d x}
      | r = \frac 1 b \paren {\frac {\d z} {\d x} - a}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac 1 b \paren {\frac {\d z} {\d x} - a}
      | r = \map f z
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = b \map f z + a
      | c = 
}}
{{end-eqn}}
This can be solved by Solution to Separable Differential Equation:
:$\ds x = \int \frac {\d z} {b \map f z + a}$
{{qed}}
Theorem: First Order ODE/y' = (x + y)^2

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 $a = b = 1$:
{{begin-eqn}}
{{eqn | l = x
      | r = \int \frac {\d z} {z^2 + 1}
      | c = 
}}
{{eqn | r = \arctan z + C_1
      | c = Primitive of $\dfrac 1 {x^2 + a^2}$
}}
{{eqn | ll= \leadsto
      | l = y - C_1
      | r = \map \arctan {x + y}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x + y
      | r = \map \tan {x + C}
      | c = letting $C = - C_1$
}}
{{end-eqn}}
{{qed}}
Theorem: First Order ODE/y' = sin^2 (x - y + 1)

The first order ODE:
:$\dfrac {\d y} {\d x} = \map {\sin^2} {x - y + 1}^2$
has the general solution:
:$\map \tan {x - y + 1} = x + C$

Proof:
Make the substitution:
:$z = x - y + 1$
Then from First Order ODE in form $y' = f (a x + b y + c)$ with $a = 1, b = - 1$:
{{begin-eqn}}
{{eqn | l = x
      | r = \int \frac {\d z} {- \sin^2 z + 1}
      | c = 
}}
{{eqn | r = \int \frac {\d z} {\cos^2 z}
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \int \sec^2 z \rd z
      | c = Secant is Reciprocal of Cosine
}}
{{eqn | r = \tan z + C_1
      | c = Primitive of $\sec^2 z$
}}
{{eqn | ll= \leadsto
      | l = \map \tan {x - y + 1}
      | r = x + C
      | c = letting $C = - C_1$
}}
{{end-eqn}}
{{qed}}
Theorem: First Order ODE in form y' = F ((a x + b y + c) over (d x + e y + f))

The first order ODE:
:$\dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$
such that:
:$ a e \ne b d$
can be solved by substituting:
:$x := z - h$
:$y := w - k$
where:
:$h = \dfrac {c e - b f} {a e - b d}$
:$k = \dfrac {a f - c d} {a e - b d}$
to obtain:
:$\dfrac {\d w} {\d z} = \map F {\dfrac {a z + b w} {d z + e w} }$
which can be solved by the technique of Solution to Homogeneous Differential Equation.

Proof:
We have:
:$\dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$
Make the substitutions:
:$x := z - h$
:$y := w - k$
We have:
:$\dfrac {\d x} {\d z} = 1$
:$\dfrac {\d y} {\d w} = 1$
Thus:
{{begin-eqn}}
{{eqn | l = \frac {\d w} {\d z}
      | r = \map F {\frac {a \paren {z - h} + b \paren {w - k} + c} {d \paren {z - h} + e \paren {w - k} + f} }
      | c = 
}}
{{eqn | r = \map F {\frac {a z + b w - a h - b k + c} {d z + e w - d h - e k + f} }
      | c = 
}}
{{end-eqn}}
In order to simplify this appropriately, we wish to reduce it to the form:
:$\dfrac {\d w} {\d z} = \map F {\dfrac {a z + b w} {d z + e w} }$
by finding values of $h$ and $k$ such that:
:$a h + b k = c$
:$d h + e k = f$
So:
{{begin-eqn}}
{{eqn | l = a h + b k
      | r = c
      | c = 
}}
{{eqn | l = d h + e k
      | r = f
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a d h + b d k
      | r = c d
      | c = 
}}
{{eqn | l = a d h + a e k
      | r = a f
      | c = 
}}
{{eqn | ll= \leadsto
      | l = k \paren {a e - b d}
      | r = a f - c d
      | c = 
}}
{{eqn | ll= \leadsto
      | l = k
      | r = \frac {a f - c d} {a e - b d}
      | c = 
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = a h + b k
      | r = c
      | c = 
}}
{{eqn | l = d h + e k
      | r = f
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a e h + b e k
      | r = c e
      | c = 
}}
{{eqn | l = b d h + b e k
      | r = b f
      | c = 
}}
{{eqn | ll= \leadsto
      | l = h \paren {a e - b d}
      | r = c e - b f
      | c = 
}}
{{eqn | ll= \leadsto
      | l = h
      | r = \frac {c e - b f} {a e - b d}
      | c = 
}}
{{end-eqn}}
We note that the above works {{iff}}:
:$a e - b d \ne 0 \implies a e \ne b d$
Thus:
:$(1): \quad \dfrac {\d w} {\d z} = \map F {\dfrac {a z + b w} {d z + e w} }$
Letting:
:$\map M {z, w} = a z + b w$
:$\map N {z, w} = d z + e w$
we see:
:$\map M {t z, t w} = a t z + b t w = t \paren {a z + b w} = t \, \map M {z, w}$
:$\map N {t z, t w} = d t z + e t w = t \paren {d z + e w} = t \, \map N {z, w}$
Thus, by definition, $(1)$ is homogeneous.
{{qed}}
Theorem: First Order ODE in form y' = F ((a x + b y + c) over (d x + e y + f)) where a e = b d/Formulation 1

The first order ODE:
:$(1): \quad \dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$
such that:
:$a e = b d$
can be solved by substituting:
:$z = a x + b y$
to obtain:
:$\dfrac {\d z} {\d x} = b \map F {\dfrac {a z + a c} {d z + a f} } + a$
which can be solved by Solution to Separable Differential Equation.

Proof:
When $a e = b d$, it is not possible to make the substitutions:
:$x := z - h$
:$y := w - k$
where:
:$h = \dfrac {c e - b f} {a e - b d}$
:$k = \dfrac {a f - c d} {a e - b d}$
and so to use the technique of First Order ODE in form $y' = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$.
So, we consider what needs to be done to make $(1)$ separable.
Let us make the substitution:
:$z = x + r y$
Consider what, if any, value of $r$ would make $(1)$ separable.
We have:
{{begin-eqn}}
{{eqn | l = a x + b y + c
      | r = a z - a r y + b y + c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = d x + e y + f
      | r = d z - d r y + \frac {b d} a y + f
      | c = noting that $e = \dfrac {b d} a$
}}
{{end-eqn}}
To make $(1)$ separable: we make:
:$b = a r$
and:
:$d r = \dfrac {b d} a$
which comes to the same thing: that $r = \dfrac b a$.
So, we can make the substitution:
:$z = a x + b y$
so:
:$\dfrac {\d z} {\d x} = a + b \dfrac {\d y} {\d x}$
which leaves us with:
{{begin-eqn}}
{{eqn | l = \frac {\d z} {\d x}
      | r = b \map F {\dfrac {z + c} {\frac d a z + f} } + a
      | c = 
}}
{{eqn | r = b \map F {\dfrac {a z + a c} {d z + a f} } + a
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: First Order ODE in form y' = F ((a x + b y + c) over (d x + e y + f)) where a e = b d/Formulation 2

The first order ODE:
:$(1): \quad \dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$
such that:
:$a e = b d$
can be solved by substituting:
:$z = d x + e y$
to obtain:
:$\dfrac {\d z} {\d x} = e \map F {\dfrac {b z + e c} {e z + e f} } + d$
which can be solved by Solution to Separable Differential Equation.

Proof:
When $a e = b d$, it is not possible to make the substitutions:
:$x := z - h$
:$y := w - k$
where:
:$h = \dfrac {c e - b f} {a e - b d}$
:$k = \dfrac {a f - c d} {a e - b d}$
and so to use the technique of First Order ODE in form $y' = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$.
So, we consider what needs to be done to make $(1)$ separable.
Let us make the substitution:
:$z = x + r y$
Consider what, if any, value of $r$ would make $(1)$ separable.
We have:
{{begin-eqn}}
{{eqn | l = d x + e y + f
      | r = d z - d r y + e y + f
}}
{{eqn | ll= \leadsto
      | l = a x + b y + c
      | r = a z - a r y + \dfrac {a e} d y + c
      | c = noting that $b = \dfrac {a e} d$
}}
{{end-eqn}}
To make $(1)$ separable: we make:
:$e = d r$
and:
:$a r = \dfrac {a e} d$
which comes to the same thing: that $r = \dfrac e d$.
So, we can make the substitution:
:$z = d x + e y$
so:
:$\dfrac {\d z} {\d x} = d + e \dfrac {\d y} {\d x}$
which leaves us with:
{{begin-eqn}}
{{eqn | l = \frac {\d z} {\d x}
      | r = e \map F {\dfrac {\frac b e z + c} {z + f} } + d
      | c = 
}}
{{eqn | r = e \map F {\dfrac {b z + e c} {e z + e f} } + d
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:First Order ODE in form y' = F ((a x + b y + c) over (d x + e y + f))
2gt0km97k0w56qzui82urbk80pqvoxd
Theorem: First Order ODE/exp y dx + (x exp y + 2 y) dy = 0

is an exact differential equation with solution:
:$x e^y + y^2 = C$

Proof:
Let:
:$\map M {x, y} = e^y$
:$\map N {x, y} = x e^y + 2 y$
Then:
{{begin-eqn}}
{{eqn | l = \dfrac {\partial M} {\partial y}
      | r = e^y
      | c = 
}}
{{eqn | l = \dfrac {\partial N} {\partial x}
      | r = e^y
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \dfrac {\partial f} {\partial x}
      | r = \map M {x, y}
      | c = 
}}
{{eqn | l = \dfrac {\partial f} {\partial y}
      | r = \map N {x, y}
      | c = 
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = f
      | r = \int \map M {x, y} \rd x + \map g y
      | c = 
}}
{{eqn | r = \int e^y \rd x + \map g y
      | c = 
}}
{{eqn | r = x e^y + \map g y
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = f
      | r = \int \map N {x, y} \rd y + \map h x
      | c = 
}}
{{eqn | r = \int \paren {x e^y + 2 y} \rd y + \map h x
      | c = 
}}
{{eqn | r = x e^y + y^2 + \map h x
      | c = 
}}
{{end-eqn}}
Thus:
:$\map f {x, y} = x e^y + y^2$
and by Solution to Exact Differential Equation, the solution to $(1)$ is:
:$x e^y + y^2 = C$
{{qed}}
Theorem: First Order ODE/(x + (2 over y)) dy + y dx = 0

is an exact differential equation with solution:
:$x y + 2 \ln y = C$

Proof:
Let:
:$\map M {x, y} = y$
:$\map N {x, y} = x + \dfrac 2 y$
Then:
{{begin-eqn}}
{{eqn | l = \dfrac {\partial M} {\partial y}
      | r = 1
      | c = 
}}
{{eqn | l = \dfrac {\partial N} {\partial x}
      | r = 1
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \dfrac {\partial f} {\partial x}
      | r = \map M {x, y}
      | c = 
}}
{{eqn | l = \dfrac {\partial f} {\partial y}
      | r = \map N {x, y}
      | c = 
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = f
      | r = \map M {x, y} \rd x + \map g y
      | c = 
}}
{{eqn | r = \int y \rd x + \map g y
      | c = 
}}
{{eqn | r = x y + \map g y
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = f
      | r = \map N {x, y} \rd y + \map h x
      | c = 
}}
{{eqn | r = \int \paren {x + \dfrac 2 y} \rd y + \map h x
      | c = 
}}
{{eqn | r = x y + 2 \ln y + \map h x
      | c = 
}}
{{end-eqn}}
Thus:
:$\map f {x, y} = x y + 2 \ln y$
and by Solution to Exact Differential Equation, the solution to $(1)$ is:
:$x y + 2 \ln y = C$
{{qed}}