avewright/proofwiki-math huggingface dataset:
Theorem: First Order ODE/x y' = Root of (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:
{{begin-eqn}}
{{eqn | l = \frac {\d x} {\d y}
      | r = \frac {\sqrt {x^2 + y^2} } x
      | c = 
}}
{{eqn | r = \sqrt {\frac {x^2 + y^2} {x^2} }
      | c = 
}}
{{eqn | r = \sqrt {1 + \paren {\frac y x}^2}
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \ln x
      | r = \int \frac {\d z} {\sqrt {1 + z^2} - z}
      | c = 
}}
{{eqn | r = \int \paren {\sqrt{1 + z^2} + z} \rd z
      | c = 
}}
{{eqn | r = \frac {z \sqrt {1 + z^2} } 2 + \frac {\map \ln {\sqrt{1 + z^2} + z} } 2 + \frac {z^2} 2 + k
      | c = Primitive of $\sqrt {x^2 + a^2}$
}}
{{eqn | ll= \leadsto
      | l = 2 \ln x
      | r = z \sqrt {1 + z^2} + \map \ln {\sqrt{1 + z^2} + z} + z^2 + C
      | c = where $C = 2 k$
}}
{{eqn | r = \frac y x \sqrt {1 + \paren {\frac y x}^2} + \map \ln {\sqrt {1 + \paren {\frac y x}^2} + \frac y x} + \paren {\frac y x}^2 + C
      | c = 
}}
{{eqn | r = \frac y x \sqrt {\frac {x^2 + y^2} {x^2} } + \map \ln {\sqrt {\frac {x^2 + y^2} {x^2} } + \frac y x} + \paren {\frac y x}^2 + C
      | c = 
}}
{{eqn | r = \frac y x \frac {\sqrt {x^2 + y^2} } x + \map \ln {\frac {\sqrt {x^2 + y^2} } x + \frac y x} + \paren {\frac y x}^2 + C
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 2 x^2 \ln x
      | r = y \sqrt {x^2 + y^2} + x^2 \map \ln {y + \sqrt {x^2 + y^2} } - x^2 \ln x + y^2 + C x^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 3 x^2 \ln x
      | r = y \sqrt {x^2 + y^2} + x^2 \map \ln {y + \sqrt {x^2 + y^2} } + y^2 + C x^2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Bernoulli's Equation/y^2 dx = (x^3 - x y) dy

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

Proof:
Dividing $(1)$ by $y^2$ and rearranging:
:$(2): \quad \dfrac {\d x} {\d y} + \dfrac x y = \dfrac {x^3} {y^2}$
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:
:$\map P y = \dfrac 1 y$
:$\map Q y = \dfrac 1 {y^2}$
:$n = 3$
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:
{{begin-eqn}}
{{eqn | l = \paren {1 - n} \int \map P y \rd y
      | r = \paren {1 - 3} \int \dfrac 1 y \rd y
      | c = 
}}
{{eqn | r = -2 \ln y
      | c = 
}}
{{eqn | r = \map \ln {y^{-2} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \mu y
      | r = e^{\map \ln {y^{-2} } }
      | c = 
}}
{{eqn | r = \dfrac 1 {y^2}
      | c = 
}}
{{end-eqn}}
and so substituting into $(3)$:
{{begin-eqn}}
{{eqn | l = \frac 1 {y^2} \frac 1 {x^2}
      | r = \paren {1 - 3} \int \frac 1 {y^2} \frac 1 {y^2} \rd y
      | c = 
}}
{{eqn | r = -2 \int \dfrac {\d y} {y^4}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac 1 {y^2 x^2}
      | r = -\dfrac 2 {-3} \dfrac 1 {y^3} + C
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 3 y
      | r = 2 x^2 + C x^2 y^2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: First Order ODE/(x^2 y^3 + y) dx = (x^3 y^2 - x) dy

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

The second order ODE:
:$y y' ' + \paren {y'}^2 - 2 y y' = 0$
has the general solution:
:$y^2 = C_2 e^{2 x} + C_1$

Proof:
Using Solution of Second Order Differential Equation with Missing Independent Variable:
{{begin-eqn}}
{{eqn | l = y p \frac {\d p} {\d y} + p^2 - 2 y p
      | r = 0
      | c = where $p = \dfrac {\d y} {\d x}$
}}
{{eqn | ll= \leadsto
      | l = \frac {\d p} {\d y} + \frac p y
      | r = 2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = p y
      | r = y^2 + C
      | c = Linear First Order ODE: $y' + \dfrac y x = k x^n$: $k = 2, n = 0$
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\d y} {\d x}
      | r = \frac {y^2 + C} y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int \dfrac {y \rd y} {y^2 + C}
      | r = \int \d x
      | c = Solution to Separable Differential Equation
}}
{{eqn | ll= \leadsto
      | l = \frac 1 2 \, \map \ln {y^2 + C}
      | r = x + k
      | c = Primitive of Function under its Derivative
}}
{{end-eqn}}
After algebra, and reassigning constants:
:$y^2 = C_2 e^{2 x} + C_1$
{{qed}}
Theorem: Rationals are Everywhere Dense in Sorgenfrey Line

$\Q$ is everywhere dense in the Sorgenfrey line.

Proof:
Let $T = \struct {\R, \tau}$ be the Sorgenfrey line.
Define:
:$\BB := \set {\hointr x y: x, y \in \R}$
where $\hointr x y$ denotes the right half-open real interval between $x$ and $y$.
By definition of Sorgenfrey line:
:$\BB$ is basis of $T$.
By definition of subset:
:$\Q^- \subseteq \R$
where $\Q^-$ denotes the topological closure of $\Q$ in $T$.
By definition of set equality to prove the equality: $\Q^- = \R$, it is necessary to show:
:$\R \subseteq \Q^-$
Let $x \in \R$.
By Characterization of Closure by Basis it suffices to prove that:
:$\forall U \in \BB: x \in U \implies U \cap \Q \ne \O$
Let $U \in \BB$.
By definition of $\BB$:
:$\exists y, z \in \R: U = \hointr y z$
Assume:
:$x \in U$
By definition of half-open real interval:
:$y \le x < z$
By Between two Real Numbers exists Rational Number:
:$\exists q \in \Q: y < q < z$
By definition of half-open real interval:
:$q \in U$
Thus by definitions of intersection and empty set:
:$U \cap \Q \ne \O$
{{qed}}
Theorem: Linear First Order ODE/x dy + y dx = x cosine x dx

The linear first order ODE:
:$x \rd y + y \rd x = x \cos x \rd x$
has the general solution:
:$x y = x \sin x + \cos x + C$

Proof:
Rearranging:
:$\dfrac {d y} {\d x} + \dfrac y x = \cos x$
This is a linear first order ODE in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where:
:$\map P x y = \dfrac 1 x$
:$\map Q x = \cos x$
Thus:
{{begin-eqn}}
{{eqn | l = \int \map P x \rd x
      | r = \int \frac 1 x \rd x
      | c = 
}}
{{eqn | r = \ln x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = e^{\int P \rd x}
      | r = e^{\ln x}
      | c = 
}}
{{eqn | r = x
      | c = 
}}
{{end-eqn}}
Thus from Solution by Integrating Factor:
{{begin-eqn}}
{{eqn | l = \dfrac {\d} {\d x} \paren {x y}
      | r = x \cos x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x y
      | r = \int x \cos x \rd x
      | c = 
}}
{{eqn | r = x \sin x + \cos x + C
      | c = Primitive of $x \cos a x$
}}
{{end-eqn}}
{{qed}}
Theorem: Linear First Order ODE/x dy + y dx = x cosine x dx

The linear first order ODE:
:$x \rd y + y \rd x = x \cos x \rd x$
has the general solution:
:$x y = x \sin x + \cos x + C$

Proof:
{{begin-eqn}}
{{eqn | l = x \dfrac {\d y} {\d x} + y
      | r = x \cos x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x y
      | r = \int x \cos x \rd x + C
      | c = Linear First Order ODE: $x y' + y = \map f x$
}}
{{eqn | r = x \sin x + \cos x + C
      | c = Primitive of $x \cos a x$
}}
{{end-eqn}}
{{qed}}
Theorem: First Order ODE/x y dy = x^2 dy + y^2 dx

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:
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:
{{begin-eqn}}
{{eqn | l = \paren {1 - n} \int \map P y \rd y
      | r = \paren {1 - 2} \int -\dfrac 1 y \rd y
      | c = 
}}
{{eqn | r = -\paren {-\ln y}
      | c = 
}}
{{eqn | r = \ln y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \mu y
      | r = e^{\ln y}
      | c = 
}}
{{eqn | r = y
      | c = 
}}
{{end-eqn}}
and so substituting into $(3)$:
{{begin-eqn}}
{{eqn | l = \frac y x
      | r = \paren {1 - 2} \int -\dfrac 1 {y^2} y \rd y
      | c = 
}}
{{eqn | r = \int \dfrac 1 y \rd y
      | c = 
}}
{{eqn | r = \ln y + C
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = x \ln y + C x
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: First Order ODE/x y dy = x^2 dy + y^2 dx

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:
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 $x, y$:
{{begin-eqn}}
{{eqn | l = \map M {t x, t y}
      | r = \paren {t y}^2
      | c = 
}}
{{eqn | r = t^2 \paren {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 - \paren {t x}^2
      | c = 
}}
{{eqn | r = t^2 \paren {x y - 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:
:$\dfrac {\d x} {\d y} = \dfrac {x^2 - x y} {y^2}$
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:
{{begin-eqn}}
{{eqn | l = \ln y
      | r = \int \frac {\d z} {\frac {z - z^2} 1 - z} + C
      | c = 
}}
{{eqn | r = -\int \frac {\d z} {z^2} + C
      | c = 
}}
{{eqn | r = \frac 1 z + C
      | c = Primitive of Power
}}
{{eqn | ll= \leadsto
      | l = \ln y
      | r = \frac y x + C
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = x \ln y + C x
      | c = multiplying through by $x$
}}
{{end-eqn}}
{{qed}}
Theorem: First Order ODE/(exp x - 3 x^2 y^2) y' + y exp x = 2 x y^3

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:
{{begin-eqn}}
{{eqn | l = \dfrac {\partial M} {\partial y}
      | r = e^x - 6 x y^2
      | c = 
}}
{{eqn | l = \dfrac {\partial N} {\partial x}
      | r = e^x - 6 x y^2
      | 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 e^x - 2 x y^3} \rd x + \map g y
      | c = 
}}
{{eqn | r = y e^x - x^2 y^3 + \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 {e^x - 3 x^2 y^2} \rd y + \map h x
      | c = 
}}
{{eqn | r = y e^x + x^2 y^3 + \map h x
      | c = 
}}
{{end-eqn}}
Thus:
:$\map f {x, y} = y e^x + x^2 y^3$
and by Solution to Exact Differential Equation, the solution to $(1)$ is:
:$y e^x + x^2 y^3 = C$
{{qed}}
Theorem: Second Order ODE/y'' + 2 x (y')^2 = 0

The second order ODE:
:$(1): \quad y' ' + 2 x \paren {y'}^2 = 0$
has the general solution:
:$C_1 \map \arctan {C_1 x} = y + C_2$

Proof:
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable.
Substitute $p$ for $y'$ in $(1)$ and rearranging:
{{begin-eqn}}
{{eqn | l = \dfrac {\d p} {\d x}
      | r = -2 x p^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d p} {p^2}
      | r = -2 \int x \rd x
      | c = Solution to Separable Differential Equation
}}
{{eqn | ll= \leadsto
      | l = -\frac 1 p
      | r = -x^2 + k^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d x} {\d y}
      | r = x^2 + k^2
      | c = substituting back for $p$
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {x^2 + k^2} 
      | r = \int \rd y
      | c = Solution to Separable Differential Equation
}}
{{eqn | ll= \leadsto
      | l = \frac 1 k \map \arctan {\frac x k}
      | r = y + C_2
      | c = Primitive of $\dfrac 1 {x^2 + a^2}$
}}
{{eqn | ll= \leadsto
      | l = C_1 \map \arctan {C_1 x}
      | r = y + C_2
      | c = setting $C_1 = \dfrac 1 k$
}}
{{end-eqn}}
{{qed}}
Theorem: First Order ODE/(y over x^2) dx + (y - 1 over x) dy = 0

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:
{{begin-eqn}}
{{eqn | l = \frac {\partial M} {\partial y}
      | r = \frac 1 {x^2}
      | c = 
}}
{{eqn | l = \dfrac {\partial N} {\partial x}
      | r = \dfrac 1 {x^2}
      | 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 \dfrac y {x^2} \rd x + \map g y
      | c = 
}}
{{eqn | r = - \frac y 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 {y - \dfrac 1 x} \r d y + \map h x
      | c = 
}}
{{eqn | r = \frac {y^2} 2 - \frac y x + \map h x
      | c = 
}}
{{end-eqn}}
Thus:
:$\map f {x, y} = \dfrac {y^2} 2 - \dfrac y x$
By Solution to Exact Differential Equation, the solution to $(1)$ is:
:$\dfrac {y^2} 2 - \dfrac y x = C$
{{qed}}
</onlyinclude>
Category:Examples of Exact Differential Equation
Category:Examples of First Order ODEs
si1zn6yfspwvxb8arsbjaq2e2j9es6u
Theorem: Focal Property of Parabola

Let $\KK$ be a parabola whose focus is at $F$ and whose directrix is $D$.
Let $P$ be an arbitrary point on $\KK$.
The tangent to $\KK$ at $P$ bisects the angle between the straight line $PF$ and the straight line through $P$ perpendicular to $D$. 
:500px

Proof:
500px
Let $BQ$ be the straight line constructed perpendicular to $D$ passing through $P$.
Let $AT$ be the tangent to $\KK$ at $P$.
Let $\angle APF$ be constructed equal to $\angle BPT$ where $F$ is on the axis of $\KK$.
It is to be shown that $F$ is the focus of $\KK$.
From Two Straight Lines make Equal Opposite Angles:
:$\angle BPT = \angle APQ$
and so as $\angle APF = \angle BPT$ by construction:
:$\angle APQ = \angle APF$
Consider the two triangles $\triangle APQ$ and $\triangle APF$.
We have that:
:$AP$ is common
:$\angle APQ = \angle APF$
From Triangle Side-Angle-Side Congruence $\triangle APQ$ and $\triangle APF$ will be congruent {{iff}} $PF = PQ$.
From the focus-directrix property of parabola $P$ is on the locus of the points that are equidistant from $F$ and $D$.
Hence the point $F$ is the focus.
{{qed}}
{{questionable|We assume the congruence of $\triangle APQ$ and $\triangle APF$, and from that assumption deduce that $F$ must be the focus. But we have not proven that $\triangle APQ$ and $\triangle APF$ are congruent, and we cannot therefore deduce that $PF {{=}} PQ$ and therefore that $F$ is the focus of $\KK$.}}
Theorem: First Order ODE/(3 x^2 over y^4 - 1 over y^2) dy - 2 x over y^3 dx = 0

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:
{{begin-eqn}}
{{eqn | l = \frac {\partial M} {\partial y}
      | r = \frac {6 x} {2 y^4}
      | c = 
}}
{{eqn | l = \dfrac {\partial N} {\partial x}
      | r = \dfrac {6 x} {y^4}
      | 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 - \dfrac {2 x} {y^3} \rd x + \map g y
      | c = 
}}
{{eqn | r = -\dfrac {x^2} {y^3} + \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 {3 x^2} {y^4} - \dfrac 1 {y^2} } \rd y + \map h x
      | c = 
}}
{{eqn | r = -\dfrac {x^2} {y^3} + \dfrac 1 y + \map h x
      | c = 
}}
{{end-eqn}}
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:
:$\dfrac 1 y - \dfrac {x^2} {y^3} = C$
{{qed}}
Category:Examples of Exact Differential Equation
Category:Examples of First Order ODEs
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Theorem: First Order ODE/(y - 1 over x) dx + (x - y) dy = 0

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:
{{begin-eqn}}
{{eqn | l = \frac {\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 - \dfrac 1 x} \rd x + \map g y
      | c = 
}}
{{eqn | r = x y - \ln 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 {x - y} \rd y + \map h x
      | c = 
}}
{{eqn | r = x y - \dfrac {y^2} 2 + \map h x
      | c = 
}}
{{end-eqn}}
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:
:$x y - \ln x - \dfrac {y^2} 2 + C$
{{qed}}
Category:Examples of Exact Differential Equation
Category:Examples of First Order ODEs
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Theorem: First Order ODE/1 over x^3 y^2 dx + (1 over x^2 y^3 + 3 y) dy = 0

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:
{{begin-eqn}}
{{eqn | l = \frac {\partial M} {\partial y}
      | r = -\frac 1 {x^3 y^2}
      | c = 
}}
{{eqn | l = \dfrac {\partial N} {\partial x}
      | r = -\frac 1 {x^3 y^2}
      | c = 
}}
{{end-eqn}}
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 \frac 1 {x^3 y^2} \rd x + \map g y
      | c = 
}}
{{eqn | r = -\frac 1 {2 x^2 y^2} + \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 {\frac 1 {x^2 y^3} + 3 y} \rd y + \map h x
      | c = 
}}
{{eqn | r = -\frac 1 {2 x^2 y^2} + 3 \dfrac {y^2} 2 + \map h x
      | c = 
}}
{{end-eqn}}
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:
:$-\dfrac 1 {2 x^2 y^2} + \dfrac {3 y^2} 2 = C$
{{qed}}
Category:Examples of Exact Differential Equation
Category:Examples of First Order ODEs
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Theorem: Linear First Order ODE/y' - (y over x) = 3 x

The linear first order ODE:
:$(1): \quad \dfrac {\d y} {\d x} - \dfrac y x = 3 x$
has the general solution:
:$\dfrac y x = 3 x + C$
or:
:$y = 3 x^2 + C x$

Proof:
This is an instance of Linear First Order ODE: $y' - \dfrac y x = k x$.
Its solution is:
:$\dfrac y x = k x + C$
or:
:$y = k x^2 + C x$
from which the solution to $(1)$ is found by substituting $3$ for $k$.
{{qed}}
Category:Examples of Linear First Order ODEs
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