avewright/proofwiki-math huggingface dataset:
Theorem: First Order ODE/(y - x^3) dx + (x + y^3) dy = 0

is an exact differential equation with solution:
:$4 x y - x^4 + y^4 = C$

Proof:
Let:
:$\map M {x, y} = y - x^3$
:$\map N {x, y} = x + y^3$
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 \paren {y - x^3} \rd x + \map g y
      | c = 
}}
{{eqn | r = x y - \dfrac {x^4} 4 + \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 \left({x + y^3}\right) \rd y + \map h x
      | c = 
}}
{{eqn | r = x y + \dfrac {y^4} 4 + \map h x
      | c = 
}}
{{end-eqn}}
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 $C = 4 C_1$:
:$4 x y - x^4 + y^4 = C$
{{qed}}
Theorem: First Order ODE/(y + y cosine x y) dx + (x + x cosine x y) dy = 0

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

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

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

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:
{{begin-eqn}}
{{eqn | l = \dfrac {\partial M} {\partial y}
      | r = -e^y + \cos x \sin y
      | c = 
}}
{{eqn | l = \dfrac {\partial N} {\partial x}
      | r = -e^y + \cos x \sin 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 = \map M {x, y} \rd x + \map g y
      | c = 
}}
{{eqn | r = -\int \paren {e^y + \cos x \cos y} \rd x + \map g y
      | c = 
}}
{{eqn | r = -x e^y - \sin x \cos 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 {\sin x \sin y - x e^y} \rd y + \map h x
      | c = 
}}
{{eqn | r = -\sin x \cos y - x e^y + \map h x
      | c = 
}}
{{end-eqn}}
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:
:$\sin x \cos y + x e^y = C$
{{qed}}
Theorem: First Order ODE/-1 over y sine x over y dx + x over y^2 sine x over y dy

is an exact differential equation with solution:
:$\dfrac x y = C$

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:
{{begin-eqn}}
{{eqn | l = \frac {\partial M} {\partial y}
      | r = \frac 1 {y^2} \sin \frac x y + \paren {-\frac 1 y} \paren {-\frac 1 {y^2} } x \cos \frac x y
      | c = 
}}
{{eqn | r = \frac 1 {y^2} \sin \frac x y + \frac x {y^3} \cos \frac x y
      | c = 
}}
{{eqn | l = \frac {\partial N} {\partial x}
      | r = \frac 1 {y^2} \sin \frac x y + \frac x {y^2} \frac 1 y \cos \frac x y
      | c = 
}}
{{eqn | r = \frac 1 {y^2} \sin \frac x y + \frac x {y^3} \cos \frac x 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 =\map M {x, y} \rd x + \map g y
      | c = 
}}
{{eqn | r = \int \paren {-\dfrac 1 y \sin \dfrac x y} \rd x + \map g y
      | c = 
}}
{{eqn | r = -\dfrac 1 y y \cos \dfrac x y + \map g y
      | c = 
}}
{{eqn | r = -\cos \dfrac x 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 {\dfrac x {y^2} \sin \dfrac x y} \rd y + \map h x
      | c = 
}}
{{end-eqn}}
Set $z = \dfrac x y$:
{{begin-eqn}}
{{eqn | l = z
      | r = \frac x y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = -\frac x {y^2}
      | c = 
}}
{{end-eqn}}
Thus by Integration by Substitution:
{{begin-eqn}}
{{eqn | l = \int \frac x {y^2} \sin \frac x y \rd y
      | r = \int \frac x {y^2} \frac {-y^2} x \sin z \rd z
      | c = 
}}
{{eqn | r = -\int \sin z \rd z
      | c = 
}}
{{eqn | r = \cos z
      | c = 
}}
{{eqn | r = \cos \frac x y
      | c = 
}}
{{end-eqn}}
Thus:
:$\map f {x, y} = -\cos \dfrac x y$
and by Solution to Exact Differential Equation, the solution to $(1)$, after simplification, is:
{{begin-eqn}}
{{eqn | l = \cos \dfrac x y
      | r = C_1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \arccos {\cos \dfrac x y}
      | r = \arccos C_1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac x y
      | r = C
      | c = where $C = \arccos C_1$
}}
{{end-eqn}}
{{qed}}
Theorem: First Order ODE/(2 x y^3 + y cosine x) dx + (3 x^2 y^2 + sine x) dy

is an exact differential equation with solution:
:$x^2 y^3 + y \sin x = C$

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:
{{begin-eqn}}
{{eqn | l = \frac {\partial M} {\partial y}
      | r = 2 x \cdot 3 y^2 + \cos x
      | c = 
}}
{{eqn | r = 6 x y^2 + \cos x
      | c = 
}}
{{eqn | l = \frac {\partial N} {\partial x}
      | r = 3 x^2 \cdot 2 y^2 + \cos x
      | c = 
}}
{{eqn | r = 6 x y^2 + \cos x
      | 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 \paren {2 x y^3 + y \cos x} \rd x + \map g y
      | c = 
}}
{{eqn | r = x^2 y^3 + y \sin x + \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 {3 x^2 y^2 + \sin x} \rd y + \map h x
      | c = 
}}
{{eqn | r = x^2 y^3 + y \sin x + \map h x
      | c = 
}}
{{end-eqn}}
Thus:
:$\map f {x, y} = x^2 y^3 + y \sin x$
and by Solution to Exact Differential Equation, the solution to $(1)$ is:
:$x^2 y^3 + y \sin x = C$
{{qed}}
Theorem: Limit Point of Subset is Limit Point of Set

Let $T = \struct {S, \tau}$ be a topological space.
Let $A, B$ be subset of $S$ such that
:$A \subseteq B$
Let $x$ be a point of $S$.
Then:
:if $x$ is limit point of $A$, then $x$ is limit point of $B$.

Proof:
Assume $x$ is limit point of $A$.
By definition of limit point it suffices to prove
:$\forall U \in \tau: x \in U \implies B \cap \paren {U \setminus \set x} \ne \O$
Let $U \in \tau$ such that
:$x \in U$
By definition of limit point:
:$A \cap \paren {U \setminus \set x} \ne \O$
By Set Intersection Preserves Subsets/Corollary:
:$A \cap \paren {U \setminus \set x} \subseteq B \cap \paren {U \setminus \set x}$
Thus:
:$B \cap \paren {U \setminus \set x} \ne \O$
{{qed}}
Category:Limit Points
9wavooswmawzbhvl8fgyr4757e2ghdf
Theorem: Existence of Infinitely Many Integrating Factors

Let the first order ordinary differential equation:
:$(1): \quad \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
be such that $M$ and $N$ are real functions of two variables which are ''not'' homogeneous functions of the same degree.
Suppose $(1)$ has an integrating factor.
Then $(1)$ has an infinite number of integrating factors

Proof:
Let $\map F f$ be any function of $f$ which is an integrating factor of $(1)$.
Then:
:$\ds \mu \map F f \paren {\map M {x, y} \rd x + \map N {x, y} \rd y} = \map F f \rd f = \map \d {\int \map F f \rd f}$
so $\mu \map F f$ is also an integrating factor.
{{qed}}
Theorem: Integrating Factor for First Order ODE/Preliminary Work

Let the first order ordinary differential equation:
:$(1): \quad \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
be non-homogeneous and not exact.
Let $\map \mu x$ be an integrating factor for $(1)$.
Let:
:$\map P {x, y} := \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$
Then:
:$\dfrac 1 \mu = \dfrac {\map P {x, y} } {N \dfrac {\partial \mu} {\partial x} - M  \dfrac {\partial \mu} {\partial y} }$

Proof:
Let us for ease of manipulation express $(1)$ in the form of differentials:
:$(2): \quad \map M {x, y} \rd x + \map N {x, y} \rd y = 0$
Let $\mu$ be an integrating factor for $(2)$.
Then, by definition:
:$\mu \, \map M {x, y} \rd x + \mu \, \map N {x, y} \rd y = 0$
is an exact differential equation.
By Solution to Exact Differential Equation:
:$\dfrac {\map \partial {\mu M} } {\partial y} = \dfrac {\map \partial {\mu N} } {\partial x}$
Evaluating this, using the Product Rule for Derivatives:
:$\mu \dfrac {\partial M} {\partial y} + M \dfrac {\partial \mu} {\partial y} = \mu \dfrac {\partial N} {\partial x} + N \dfrac {\partial \mu} {\partial x}$
which leads us to:
:$\dfrac 1 \mu \paren {N \dfrac {\partial \mu} {\partial x} - M  \dfrac {\partial \mu} {\partial y} } = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$
Let us use $\map P {x, y}$ for $\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$.
Then:
:$(3): \quad \dfrac 1 \mu = \dfrac {\map P {x, y} } {N \dfrac {\partial \mu} {\partial x} - M \dfrac {\partial \mu} {\partial y}}$
Theorem: Integrating Factor for First Order ODE/Function of One Variable

Suppose that:
:$\map g x = \dfrac {\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x} } {\map N {x, y} }$
is a function of $x$ only.
Then:
:$\map \mu x = e^{\int \map g x \rd x}$
is an integrating factor for $(1)$.
Similarly, suppose that:
:$\map h y = \dfrac {\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x} } {\map M {x, y} }$
is a function of $y$ only.
Then:
:$\map \mu y = e^{\int -\map h y \rd y}$
is an integrating factor for $(1)$.

Proof:
=== Preliminary Work ===
{{:Integrating Factor for First Order ODE/Preliminary Work}}
Theorem: Integrating Factor for First Order ODE/Function of Sum of Variables

Suppose that:
:$\map g z = \dfrac {\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x} } {\map N {x, y} - \map M {x, y} }$
is a function of $z$, where $z = x + y$.
Then:
:$\map \mu {x + y} = \map \mu z = e^{\int \map g z \rd z}$
is an integrating factor for $(1)$.

Proof:
=== Preliminary Work ===
{{:Integrating Factor for First Order ODE/Preliminary Work}}
Theorem: Integrating Factor for First Order ODE/Function of Product of Variables

:$\map g z = \dfrac {\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}} {N y - M x}$
is a function of $z$, where $z = x y$.
Then:
:$\map \mu {x y} = \map \mu z = e^{\int \map g z \d z}$
is an integrating factor for $(1)$.

Proof:
=== Preliminary Work ===
{{:Integrating Factor for First Order ODE/Preliminary Work}}
Theorem: Set of Condensation Points of Union is Union of Sets of Condensation Points

Let $T = \left({S, \tau}\right)$ be a topological space.
Let $A, B$ be subsets of $S$.
Then:
:$\left({A \cup B}\right)^0 = A^0 \cup B^0$

Proof:
=== Set of Condensation Points of Union Subset Union of Sets of Condensation Points ===
Let $x \in \left({A \cup B}\right)^0$.
By definition of set of condensation points:
:$x$ is condensation point of $A \cup B$
By Lemma:
:$x$ is condensation point of $A$ or $x$ is condensation point of $B$
By definition of set of condensation points:
:$x \in A^0$ or $x \in B^0$
Thus by definition of union:
:$x \in A^0 \cup B^0$
{{qed|lemma}}
Theorem: Set of Condensation Points of Union is Union of Sets of Condensation Points/Lemma

Let $x$ be a point of $S$.
Then:
:if $x$ is condensation point of $A \cup B$,
:then $x$ is condensation point of $A$ or $x$ is condensation point of $B$.

Proof:
Assume $x$ is condensation point of $A \cup B$.
{{AimForCont}}
:$x$ is not condensation point of $A$ and $x$ is not condensation point of $B$.
By definition of condensation point:
:$\exists U_1 \in \tau: x \in U_1 \land A \cap U_1$ is countable
By definition of condensation point:
:$\exists U_2 \in \tau: x \in U_2 \land B \cap U_2$ is countable
By definition of intersection:
:$x \in U_1 \cap U_2$
By definition of topological space:
:$U_1 \cap U_2 \in \tau$
By Intersection is Subset:
:$A \cap U_1 \cap U_2 \subseteq A \cap U_1 \land B \cap U_1 \cap U_2 \subseteq B \cap U_2$
By Set Union Preserves Subsets:
:$\left({A \cap U_1 \cap U_2}\right) \cup \left({B \cap U_1 \cap U_2}\right) \subseteq \left({A \cap U_1}\right) \cup \left({B \cap U_2}\right)$
By Countable Union of Countable Sets is Countable:
:$\left({A \cap U_1}\right) \cup \left({B \cap U_2}\right)$ is countable
Then by Subset of Countable Set is Countable:
:$(1): \quad \left({A \cap U_1 \cap U_2}\right) \cup \left({B \cap U_1 \cap U_2}\right)$ is countable
By Intersection Distributes over Union:
:$\left({A \cap U_1 \cap U_2}\right) \cup \left({B \cap U_1 \cap U_2}\right) = \left({A \cup B}\right) \cap \left({U_1 \cap U_2}\right)$
By definition of condensation point:
:$\left({A \cup B}\right) \cap \left({U_1 \cap U_2}\right)$ is uncountable
This contradicts $(1)$.
Thus the result fallows by Proof by Contradiction.
{{qed}}
Theorem: Integrating Factor for First Order ODE/Conclusion

Let the first order ordinary differential equation:
:$(1): \quad \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
be non-homogeneous and not exact.
Let $\map \mu {x, y}$be an integrating factor for $(1)$.
If one of these is the case:
:$\mu$ is a function of $x$ only
:$\mu$ is a function of $y$ only
:$\mu$ is a function of $x + y$
:$\mu$ is a function of $x y$
then:
:$\mu = e^{\int \map f w \rd w}$
where $w$ depends on the nature of $\mu$.

Proof:
We have an equation of the form:
:$\dfrac 1 \mu \dfrac {\d \mu} {\d w} = \map f w$
which is what you get when you apply the Chain Rule for Derivatives and Derivative of Logarithm Function to:
:$\dfrac {\map \d {\ln \mu} } {\d w} = \map f w$
Thus:
:$\ds \ln \mu = \int \map f w \rd w$
and so:
:$\mu = e^{\int \map f w \rd w}$
Hence the results as stated.
{{qed}}
Theorem: First Order ODE/y dx + (x^2 y - x) dy = 0

The first order ODE:
:$(1): \quad y \rd x + \paren {x^2 y - x} \rd y = 0$
has the general solution:
:$\dfrac {y^2} 2 - \dfrac y x = C$

Proof:
We note that $(1)$ is in the form:
:$\map M {x, y} \d x + \map N {x, y} \d y = 0$
but that $(1)$ is not exact.
So, let:
:$\map M {x, y} = y$
:$\map N {x, y} = x^2 y - x$
Let:
:$\map P {x, y} = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$
Thus:
{{begin-eqn}}
{{eqn | l = \map P {x, y}
      | r = 1 - \paren {2 x y - 1}
      | c = 
}}
{{eqn | r = 2 \paren {1 - x y}
      | c = 
}}
{{end-eqn}}
It can be observed that:
{{begin-eqn}}
{{eqn | l = \frac {\map P {x, y} } {\map N {x, y} }
      | r = \frac {2 \paren {1 - x y} } {x^2 y - x}
      | c = 
}}
{{eqn | r = \frac {-2 \paren {x y - 1} } {x \paren {x y - 1} }
      | c = 
}}
{{eqn | r = -\frac 2 x
      | c = 
}}
{{end-eqn}}
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:
:$\map \mu x = e^{\int \map g x \rd x}$
Hence:
{{begin-eqn}}
{{eqn | l = \int \map g x \rd x
      | r = \int - \paren {2 / x} \rd x
      | c = 
}}
{{eqn | r = - 2 \ln x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = e^{\int \map g x \rd x}
      | r = \frac 1 {x^2}
      | c = 
}}
{{end-eqn}}
Thus an integrating factor for $(1)$ has been found:
:$\mu = \dfrac 1 {x^2}$
which yields, when multiplying it throughout $(1)$:
:$\dfrac y {x^2} \rd x + \paren {y - \dfrac 1 x} \rd y = 0$
which is now exact.
By First Order ODE: $\dfrac y {x^2} \rd x + \paren {y - \dfrac 1 x} \rd y = 0$, its solution is:
:$\dfrac {y^2} 2 - \dfrac y x = C$
{{qed}}
Theorem: First Order ODE/y dx + (x^2 y - x) dy = 0

The first order ODE:
:$(1): \quad y \rd x + \paren {x^2 y - x} \rd y = 0$
has the general solution:
:$\dfrac {y^2} 2 - \dfrac y x = C$

Proof:
Rearranging, we have:
:$x^2 y \rd y - \paren {x \rd y - y \rd x} = 0$
Aiming to use Quotient Rule for Differentials, divide by $x^2$:
:$y \rd y = \dfrac {x \rd y - y \rd x} {x^2}$
So from Quotient Rule for Differentials: Formulation 1
:$y \rd y = \map \d {\dfrac y x}$
from which the solution immediately drops:
:$\dfrac {y^2} 2 - \dfrac y x = C$
{{qed}}
Theorem: Set of Condensation Points of Countable Set is Empty

Let $T = \struct {S, \tau}$ be a topological space.
Let $A$ be a subset of $S$.
Then:
:if $A$ is countable,
:then $A^0 = \O$.

Proof:
Assume
:$A$ is countable.
{{AimForCont}}
:$A^0 \ne \O$
By definition of empty set:
:$\exists x: x \in A^0$
Then by definition of set of condensation points:
:$x$ is a condensation point of $A$.
This contradicts Lemma.
Thus the result follows by Proof by Contradiction.
{{qed}}
Theorem: Set of Condensation Points of Countable Set is Empty/Lemma

:if $A$ is countable,
:then there exists no point $x$ of $S$ such that $x$ is a condensation point of $A$.

Proof:
Assume
:$A$ is countable.
{{AimForCont}} there exists a point $x$ of $S$ such that
: $x$ is a condensation point of $A$
By definition of topological space:
:$S \in \tau$
Then by definition of condensation point:
:$A \cap S$ is uncountable
By Intersection with Subset is Subset:
:$A \cap S = A$
$A$ is countable contradicts $A$ is uncountable.
Thus the result follows by Proof by Contradiction.
{{qed}}
Theorem: First Order ODE/(3 x^2 - y^2) dy - 2 x y dx = 0

The first order ODE:
:$(1): \quad \paren {3 x^2 - y^2} \rd y - 2 x y \rd x = 0$
has the general solution:
:$\dfrac 1 y - \dfrac {x^2} {y^3} = C$

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:
{{begin-eqn}}
{{eqn | l = \map P {x, y}
      | r = -2 x - 6 x
      | c = 
}}
{{eqn | r = -8 x
      | c = 
}}
{{end-eqn}}
It can be observed that:
{{begin-eqn}}
{{eqn | l = \frac {\map P {x, y} } {\map M {x, y} }
      | r = \frac {-8 x} {-2 x y}
      | c = 
}}
{{eqn | r = \frac 4 y
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \int -\map h y \rd y
      | r = \int -\paren {4 / y} \rd y
      | c = 
}}
{{eqn | r = - 4 \ln y
      | c = 
}}
{{eqn | r = \map \ln {y^{-4} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = e^{\int -\map h y \rd y}
      | r = \frac 1 {y^4}
      | c = 
}}
{{end-eqn}}
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:
:$\dfrac 1 y - \dfrac {x^2} {y^3} = C$
{{qed}}
Theorem: First Order ODE/(x y - 1) dx + (x^2 - x y) dy = 0

The first order ODE:
:$(1): \quad \paren {x y - 1} \rd x + \paren {x^2 - x y} \rd y = 0$
has the general solution:
:$x y - \ln x - \dfrac {y^2} 2 + C$

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:
{{begin-eqn}}
{{eqn | l = \map P {x, y}
      | r = x - \paren {2 x - y}
      | c = 
}}
{{eqn | r = -\paren {x - y}
      | c = 
}}
{{end-eqn}}
It can be observed that:
{{begin-eqn}}
{{eqn | l = \frac {\map P {x, y} } {\map N {x, y} }
      | r = \frac {-\paren {x - y} } {x \paren {x - y} }
      | c = 
}}
{{eqn | r = -\frac 1 x
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \int \map g x \rd x
      | r = \int -\paren {1 / x} \rd x
      | c = 
}}
{{eqn | r = -\ln x
      | c = 
}}
{{eqn | r = \map \ln {x^{-1} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = e^{\int \map g x \rd x}
      | r = \frac 1 x
      | c = 
}}
{{end-eqn}}
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:
:$x y - \ln x - \dfrac {y^2} 2 + C$
{{qed}}