avewright/proofwiki-math huggingface dataset:
Theorem: Cycloid has Tautochrone Property

Let a wire $AB$ be curved into the shape of an arc of a cycloid such that:
:$A$ is at the cusp
:$B$ is the highest point of the arc
and inverted so that its cusps are uppermost and on the same horizontal line.
Thus $B$ is the lowest point of the arc.
Let $AB$ be embedded in a constant and uniform gravitational field where Acceleration Due to Gravity is $g$.
Let a bead $P$ be released from anywhere on the wire between $A$ and $B$ to slide down without friction to $B$.
Then the time taken for $P$ to slide to $B$ is:
:$T = \pi \sqrt {\dfrac a g}$
independently of the point from which $P$ is released.
That is, a cycloid is a tautochrone.

Proof:
:500px
By the Principle of Conservation of Energy, the speed of the bead at a particular height is determined by its loss in potential energy in getting there.
Thus, at the point $\tuple {x, y}$, we have:
:$(1): \quad v = \dfrac {\d s} {\d t} = \sqrt {2 g y}$
This can be written:
{{begin-eqn}}
{{eqn | l = \d t
      | r = \frac {\d s} {\sqrt {2 g y} }
      | c = 
}}
{{eqn | r = \frac {\sqrt {\d x^2 + \d y^2} } {\sqrt {2 g y} }
      | c = 
}}
{{end-eqn}}
Thus the time taken for the bead to slide down the wire is given by:
:$\ds T_1 = \int \sqrt {\dfrac {\d x^2 + \d y^2} {2 g y} }$
From Equation of Cycloid, we have:
:$x = a \paren {\theta - \sin \theta}$
:$y = a \paren {1 - \cos \theta}$
Substituting these in the above integral:
:$\ds T_1 = \int_0^{\theta_1} \sqrt {\dfrac {2 a^2 \paren {1 - \cos \theta} } {2 a g \paren {1 - \cos \theta} } } \rd \theta = \theta_1 \sqrt {\dfrac a g}$
This is the time needed for the bead to reach the bottom when released when $\theta_1 = \pi$, and so:
:$T_1 = \pi \sqrt {\dfrac a g}$
Now suppose the bead is released at any intermediate point $\tuple {x_0, y_0}$.
Take equation $(1)$ and replace it with:
:$v = \dfrac {\d s} {\d t} = \sqrt {2 g \paren {y - y_0} }$
Thus the total time to reach the bottom is:
{{begin-eqn}}
{{eqn | l = T
      | r = \int_{\theta_0}^\pi \sqrt {\dfrac {2 a^2 \paren {1 - \cos \theta} } {2 a g \paren {\cos \theta_0 - \cos \theta} } } \rd \theta
      | c = 
}}
{{eqn | r = \sqrt {\dfrac a g} \int_{\theta_0}^\pi \sqrt {\dfrac {1 - \cos \theta} {\cos \theta_0 - \cos \theta} } \rd \theta
      | c = 
}}
{{eqn | n = 2
      | r = \sqrt {\dfrac a g} \int_{\theta_0}^\pi \dfrac {\sin \frac 1 2 \theta \rd \theta} {\sqrt {\cos^2 \frac 1 2 \theta_0 - \cos^2 \frac 1 2 \theta} } \rd \theta
      | c = Half Angle Formula for Cosine and Half Angle Formula for Sine
}}
{{end-eqn}}
Setting:
:$u = \dfrac {\cos \frac 1 2 \theta} {\cos \frac 1 2 \theta_0}$
and so:
:$\d u = -\dfrac 1 2 \dfrac {\sin \frac 1 2 \theta \rd \theta} {\cos \frac 1 2 \theta_0}$
Then $(2)$ becomes:
{{begin-eqn}}
{{eqn | l = T
      | r = -2 \sqrt {\dfrac a g} \int_1^0 \frac {\d u} {\sqrt {1 - u^2} }
      | c = 
}}
{{eqn | r = 2 \sqrt {\dfrac a g} \Big [{\arcsin u}\Big]_0^1
      | c = 
}}
{{eqn | r = \pi \sqrt {\dfrac a g}
      | c = 
}}
{{end-eqn}}
That is, wherever the bead is released from, it takes that same time to reach the bottom.
Hence the result.
{{qed}}
Theorem: Velocity of Bead on Brachistochrone

Consider a wire bent into the shape of an arc of a cycloid $C$ and inverted so that its cusps are uppermost and on the same horizontal line.
Let $C$ be defined by Equation of Cycloid embedded in a cartesian plane:
:$x = a \paren {\theta - \sin \theta}$
:$y = a \paren {1 - \cos \theta}$
Let a bead $B$ be released from some point on the wire.
Let $B$ slide without friction under the influence of a constant gravitational field exerting an acceleration $g$.
Let $s_0$ be the arc length along the cycloid.
Let $s$ be the arc length along the cycloid at any subsequent point in time.
Then the speed $v$ of $B$ relative to $C$ is defined by the equation:
:$4 a v^2 = g \paren {2 s s_0 - s^2}$

Proof:
By Brachistochrone is Cycloid, $C$ is a brachistochrone.
{{ProofWanted}}
Theorem: Open Ball is Subset of Open Ball

Let $M = \struct {A, d}$ be a metric space.
Let $x, y$ be points of $A$.
Then:
:$\epsilon-\delta \ge \map d {x, y} \implies \map {B_\delta} y \subseteq \map {B_\epsilon} x$
where $\map {B_\epsilon} x$ denotes the open $\epsilon$-ball in $M = \struct {A, d}$.

Proof:
Let $\epsilon - \delta \ge \map d {x, y}$.
Then $\epsilon \ge \map d {x, y} + \delta$.
If $z \in \map {B_\delta} y$, then $\map d {y, z} < \delta$.
So:
{{begin-eqn}}
{{eqn | l = \map d {x, z}
      | o = \le
      | r = \map d {x, y} + \map d {y, z}
      | c = {{Metric-space-axiom|2}}
}}
{{eqn | o = <
      | r = \map d {x, y} + \delta
      | c = {{Defof|Open Ball of Metric Space}}
}}
{{eqn | o = \le
      | r = \epsilon
}}
{{end-eqn}}
Thus $z \in \map {B_\epsilon} x$.
So $\map {B_\delta} y \subseteq \map {B_\epsilon} x$.
{{qed}}
Theorem: Pursuit Curve of Boat in River

Consider a straight river $R$ whose parallel banks are aligned with the $y$-axis and the line $x = c$ of a cartesian plane.
Let the current of $R$ have a constant and uniform speed $a$ in the negative $y$ direction.
Let a boat $B$ be launched from the point $\tuple {c, 0}$ and headed directly towards the origin with speed $b$ relative to the water.
The path of $B$ is defined by the equation:
:$c^k \paren {y + \sqrt {x^2 + x^2} } = x^{k + 1}$
where $k = \dfrac a b$.

Proof:
The components of the velocity of $b$ are:
:$\dfrac {\d x} {\d t} = - b \cos \theta$
:$\dfrac {\d y} {\d t} = - a + b \sin \theta$
Hence:
{{begin-eqn}}
{{eqn | l = \frac {\d y} {\d x}
      | r = \frac {-a + b \sin \theta} {-b \cos \theta}
      | c = 
}}
{{eqn | r = \frac {-a + b \paren {-\dfrac y {\sqrt {x^2 + y^2} } } } {-b \paren {\dfrac x {\sqrt {x^2 + y^2} } } }
      | c = 
}}
{{eqn | n = 1
      | r = \frac {-a \sqrt {x^2 + y^2} + b y} {b x}
      | c = 
}}
{{end-eqn}}
$(1)$ is a homogeneous differential equation.
Let $z = \dfrac y x$.
Then by Solution to Homogeneous Differential Equation:
{{begin-eqn}}
{{eqn | l = \ln x
      | r = \int \frac {\d z} {\dfrac {-a \sqrt {1 + z^2} + b z} b - z} + C
      | c = 
}}
{{eqn | r = \int \frac {b \rd z} {-a \sqrt {1 + z^2} + b z - b z} + C
      | c = 
}}
{{eqn | r = \int \frac {b \rd z} {-a \sqrt {1 + z^2} } + C
      | c = 
}}
{{eqn | r = -\frac b a \int \frac {\d z} {\sqrt {1 + z^2} } + C
      | c = 
}}
{{eqn | r = -\frac b a \map \ln {z + \sqrt {1 + z^2} }  + C
      | c = Primitive of $\dfrac 1 {\sqrt{x^2 + a^2} }$
}}
{{end-eqn}}
Substituting back for $z$, putting $k = \dfrac a b$, and rearranging:
:$C^k \paren {y + \sqrt {x^2 + x^2} } = x^{k + 1}$
{{qed}}
Theorem: Current in Electric Circuit/L, R, C in Series

Consider the electronic circuit $K$ consisting of:
:a resistance $R$
:an inductance $L$
:a capacitance $C$
in series with a source of electromotive force $E$ which is a function of time $t$.
:File:CircuitRLCseries.png
The electric current $I$ in $K$ is given by the linear second order ODE:
:$L \dfrac {\d^2 I} {\d t^2} + R \dfrac {\d I} {\d t} + \dfrac 1 C I = \dfrac {\d E} {\d t}$

Proof:
Let:
:$E_L$ be the drop in electromotive force across $L$
:$E_R$ be the drop in electromotive force across $R$
:$E_C$ be the drop in electromotive force across $C$.
From Kirchhoff's Voltage Law:
:$E - E_L - E_R - E_C = 0$
From Ohm's Law:
:$E_R = R I$
From Drop in EMF caused by Inductance is proportional to Rate of Change of Current:
:$E_L = L \dfrac {\d I} {\d t}$
From Drop in EMF caused by Capacitance is proportional to Accumulated Charge:
:$E_C = \dfrac 1 C Q$
where $Q$ is the electric charge $Q$ that has accumulated on $C$.
Thus:
:$E - L \dfrac {\d I} {\d t} - R I - \dfrac 1 C Q = 0$
which can be rewritten:
:$L \dfrac {\d I} {\d t} + R I + \dfrac 1 C Q = E$
Differentiating {{WRT|Differentiation}} $t$ gives:
:$L \dfrac {\d^2 I} {\d t^2} + R \dfrac {\d I} {\d t} + \dfrac 1 C I = \dfrac {\d E} {\d t}$
{{qed}}
Theorem: Electric Charge in Electric Circuit/L, R, C in Series

Consider the electronic circuit $K$ consisting of:
:a resistance $R$
:an inductance $L$
:a capacitance $C$
in series with a source of electromotive force $E$ which is a function of time $t$.
:File:CircuitRLCseries.png
The electric charge $Q$ in $K$ is given by the linear second order ODE:
:$L \dfrac {\d^2 Q} {\d t^2} + R \dfrac {\d Q} {\d t} + \dfrac 1 C Q = E$

Proof:
Let:
:$E_L$ be the drop in electromotive force across $L$
:$E_R$ be the drop in electromotive force across $R$
:$E_C$ be the drop in electromotive force across $C$.
From Kirchhoff's Voltage Law:
:$E - E_L - E_R - E_C = 0$
From Ohm's Law:
:$E_R = R I$
From Drop in EMF caused by Inductance is proportional to Rate of Change of Current:
:$E_L = L \dfrac {\d I} {\d t}$
From Drop in EMF caused by Capacitance is proportional to Accumulated Charge:
:$E_C = \dfrac 1 C Q$
where $Q$ is the electric charge $Q$ that has accumulated on $C$.
Thus:
:$E - L \dfrac {\d I} {\d t} - R I - \dfrac 1 C Q = 0$
which can be rewritten:
:$L \dfrac {\d I} {\d t} + R I + \dfrac 1 C Q = E$
By definition of electric current:
:$I = \dfrac {\d Q} {\d t}$
and so:
:$L \dfrac {\d^2 Q} {\d t^2} + R \dfrac {\d Q} {\d t} + \dfrac 1 C Q = E$
{{qed}}
Theorem: Current in Electric Circuit/L, R in Series

Consider the electronic circuit $K$ consisting of:
:a resistance $R$
:an inductance $L$
in series with a source of electromotive force $E$ which is a function of time $t$.
:File:CircuitRLseries.png
The electric current $I$ in $K$ is given by the linear first order ODE:
:$L \dfrac {\d I} {\d t} + R I = E$

Proof:
Let:
:$E_L$ be the drop in electromotive force across $L$
:$E_R$ be the drop in electromotive force across $R$
From Kirchhoff's Voltage Law:
:$E - E_L - E_R = 0$
From Ohm's Law:
:$E_R = R I$
From Drop in EMF caused by Inductance is proportional to Rate of Change of Current:
:$E_L = L \dfrac {\d I} {\d t}$
Thus:
:$E - L \dfrac {\d I} {\d t} - R I = 0$
which can be rewritten:
:$L \dfrac {\d I} {\d t} + R I = E$
{{qed}}
Theorem: Current in Electric Circuit/L, R in Series/Constant EMF at t = 0

Let the electric current flowing in $K$ at time $t = 0$ be $I_0$.
Let a constant EMF $E_0$ be imposed upon $K$ at time $t = 0$.
The electric current $I$ in $K$ is given by the equation:
:$I = \dfrac {E_0} R + \paren {I_0 - \dfrac {E_0} R} e^{-R t / L}$

Proof:
From Electric Current in Electric Circuit: L, R in Series:
:$L \dfrac {\d I} {\d t} + R I = E_0$
defines the behaviour of $I$.
This can be written as:
:$(1): \quad \dfrac {\d I} {\d t} = \dfrac R L \paren {\dfrac {E_0} R - I}$
$(1)$ is in the form:
:$\dfrac {\d y} {\d x} = k \paren {y_a - y}$
where:
:$k \in \R: k > 0$
:$y = y_0$ at $x = 0$
This is an example of the Decay Equation, where:
:$k = \dfrac R L$
:$y_a = \dfrac {E_0} R$
:$y_0 = I_0$.
whose particular solution is:
:$y = y_a + \paren {y_0 - y_a} e^{-k x}$
Hence the particular solution to $(1)$ is:
:$I = \dfrac {E_0} R + \paren {I_0 - \dfrac {E_0} R} e^{-R t / L}$
{{qed}}
Theorem: Current in Electric Circuit/L, R in Series/Constant EMF at t = 0/Corollary 1

Let the electric current flowing in $K$ at time $t = 0$ be $I_0$.
Let a constant EMF $E_0$ be imposed upon $K$ at time $t = 0$.
After a sufficiently long time, the electric current $I$ in $K$ is given by the equation:
:$E_0  = R I$

Proof:
From Electric Current in Electric Circuit: L, R in Series: Constant EMF at $t = 0$:
:$I = \dfrac {E_0} R + \paren {I_0 - \dfrac {E_0} R} e^{-R t / L}$
We have that:
:$\ds \lim_{t \mathop \to \infty} e^{-R t / L} \to 0$
and so:
:$\ds \lim_{t \mathop \to \infty} I \to \dfrac {E_0} R$
Hence the result.
{{qed}}
Theorem: Current in Electric Circuit/L, R in Series/Constant EMF at t = 0/Corollary 2

Let the electric current flowing in $K$ at time $t = 0$ be $0$.
Let a constant EMF $E_0$ be imposed upon $K$ at time $t = 0$.
The electric current $I$ in $K$ is given by the equation:
:$I = \dfrac {E_0} R \paren {1 - e^{-R t / L} }$

Proof:
From Electric Current in Electric Circuit: L, R in Series: Constant EMF at $t = 0$:
:$I = \dfrac {E_0} R + \paren {I_0 - \dfrac {E_0} R} e^{-R t / L}$
Setting $I_0 = 0$:
{{begin-eqn}}
{{eqn | l = I
      | r = \dfrac {E_0} R + \paren {0 - \dfrac {E_0} R} e^{-R t / L}
      | c = 
}}
{{eqn | r = \dfrac {E_0} R \paren {1 - e^{-R t / L} }
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Current in Electric Circuit/L, R in Series/Constant EMF at t = 0/Corollary 3

Let the electric current flowing in $K$ at time $t = 0$ be $I_0$.
Let  EMF imposed upon $K$ be zero.
The electric current $I$ in $K$ is given by the equation:
:$I = I_0 e^{-R t / L}$

Proof:
From Electric Current in Electric Circuit: L, R in Series: Constant EMF at $t = 0$:
:$I = \dfrac {E_0} R + \paren {I_0 - \dfrac {E_0} R} e^{-R t / L}$
Setting $E_0 = 0$:
{{begin-eqn}}
{{eqn | l = I
      | r = \dfrac 0 R + \paren {I_0 - \dfrac 0 R} e^{-R t / L}
      | c = 
}}
{{eqn | r = I_0 e^{-R t / L}
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Current in Electric Circuit/L, R in Series/Exponentially Decaying EMF at t = 0

Let the electric current flowing in $K$ at time $t = 0$ be $I_0$.
Let an EMF $E$ be imposed upon $K$ at time $t = 0$ defined by the equation:
:$E = E_0 e^{-k t}$
The electric current $I$ in $K$ is given by the equation:
:$I = \dfrac {E_0} {R - k L} e^{-k t} + \paren {I_0 - \dfrac {E_0} {R - k L} } e^{-R t / L}$

Proof:
From Electric Current in Electric Circuit: L, R in Series:
:$L \dfrac {\d I} {\d t} + R I = E_0 e^{-k t}$
defines the behaviour of $I$.
This can be written as:
:$(1): \quad \dfrac {\d I} {\d t} + \dfrac R L I = \dfrac {E_0} L  e^{-k t}$
$(1)$ is a linear first order ODE in the form:
:$\dfrac {\d I} {\d t} + \map P t I = \map Q t$
where:
:$\map P t = \dfrac R L$
:$\map Q t = \dfrac {E_0} L  e^{-k t}$
Thus:
{{begin-eqn}}
{{eqn | l = \int \map P t \rd t
      | r = \int \dfrac R L \rd t
      | c = 
}}
{{eqn | r = \dfrac {R t} L
      | c = 
}}
{{eqn | ll= \leadsto
      | l = e^{\int P \rd t}
      | r = e^{R t / L}
      | c = 
}}
{{end-eqn}}
Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
:$\map {\dfrac \d {\d t} } {e^{R t / L} I} = e^{R t / L} \dfrac {E_0} L e^{-k t}$
and the general solution becomes:
:$\ds e^{R t / L} I = \int e^{R t / L} \dfrac {E_0} L e^{-k t} \rd t$
and so:
{{begin-eqn}}
{{eqn | l = e^{R t / L} I
      | r = \frac {E_0} L  \int e^{R t / L} e^{-k t} \rd t
      | c = 
}}
{{eqn | r = \frac {E_0} L  \int e^{\paren {R / L - k} t} \rd t
      | c = 
}}
{{eqn | r = \frac {E_0} {L \paren {\frac R L - k} } e^{\paren {R / L - k} t} + C
      | c = 
}}
{{eqn | r = \frac {E_0} {R - k L} e^{\paren {R / L - k} t} + C
      | c = 
}}
{{end-eqn}}
When $t = 0$, we have $I = I_0$.
So:
{{begin-eqn}}
{{eqn | l = I_0
      | r = \frac {E_0} {R - k L} + C
      | c = 
}}
{{eqn | ll= \leadsto
      | l = C
      | r = I_0 - \frac {E_0} {R - k L}
      | c = 
}}
{{end-eqn}}
So:
:$I e^{R t/ L} = \dfrac {E_0} {R - k L} e^{\paren {R / L - k} t} + I_0 - \dfrac {E_0} {R - k L}$
Multiplying by $e^{\frac {R t} L}$ and tidying up, we get:
{{begin-eqn}}
{{eqn | l = I
      | r = \frac {E_0} {R - k L} e^{- R t / L + R t / L - k t} + I_0 e^{-R t / L} - \dfrac {E_0} {R - k L} e^{-R t / L}
      | c = 
}}
{{eqn | r = \frac {E_0} {R - k L} e^{- k t} + I_0 e^{-R t / L} - \frac {E_0} {R - k L} e^{-R t / L}
      | c = 
}}
{{eqn | r = \frac {E_0} {R - k L} e^{- k t} + \paren {I_0 - \frac {E_0} {R - k L} } e^{-R t / L}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Current in Electric Circuit/L, R in Series/Sinusoidal EMF

Let the electric current flowing in $K$ at time $t = 0$ be $I_0$.
Let an EMF $E$ be imposed upon $K$ at time $t = 0$ defined by the equation:
:$E = E_0 \sin \omega t$
The electric current $I$ in $K$ is given by the equation:
:$I = \dfrac {E_0} {\sqrt {R^2 - L^2 \omega^2} } \map \sin {\omega t - \alpha} + \paren {I_0 - \dfrac {E_0 L \omega} {R^2 + L^2 \omega^2} } e^{-R t / L}$
where $\tan \alpha = \dfrac {L \omega} R$.

Proof:
From Electric Current in Electric Circuit: L, R in Series:
:$L \dfrac {\d I} {\d t} + R I = E_0 \sin \omega t$
defines the behaviour of $I$.
This can be written as:
:$(1): \quad \dfrac {\d I} {\d t} + \dfrac R L I = \dfrac {E_0} L \sin \omega t$
$(1)$ is a linear first order ODE in the form:
:$\dfrac {\d I} {\d t} + \map P t I = \map Q t$
where:
:$\map P t = \dfrac R L$
:$\map Q t = \dfrac {E_0} L  e^{-k t}$
Thus:
{{begin-eqn}}
{{eqn | l = \int \map P t \rd t
      | r = \int \dfrac R L \rd t
      | c = 
}}
{{eqn | r = \dfrac {R t} L
      | c = 
}}
{{eqn | ll= \leadsto
      | l = e^{\int P \rd t}
      | r = e^{R t / L}
      | c = 
}}
{{end-eqn}}
Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
:$\map {\dfrac \d {\d t} } {e^{R t / L} I} = e^{R t / L} \dfrac {E_0} L \sin \omega t$
and so the general solution becomes:
{{begin-eqn}}
{{eqn | l = e^{R t / L} I
      | r = \frac {E_0} L \int e^{R t / L} \sin \omega t \rd t
      | c = 
}}
{{eqn | r = \frac {E_0} L \frac {e^{R t / L} \paren {\frac R L \sin \omega t - \omega \cos \omega t} } {\paren {\frac R L}^2 + \omega^2} + C
      | c = Primitive of $e^{a x} \sin b x$
}}
{{eqn | r = \frac {E_0 \paren {R \sin \omega t - L \omega \cos \omega t} } {R^2 + \omega^2 L^2} e^{R t / L} + C
      | c = 
}}
{{end-eqn}}
When $t = 0, I = I_0$.
{{begin-eqn}}
{{eqn | l = I_0
      | r = E_0 \, {\frac {-L \omega} {R^2 + \omega^2 L^2} } + C
      | c = 
}}
{{eqn | ll= \leadsto
      | l = C
      | r = I_0 + \frac {E_0 L \omega} {R^2 + \omega^2 L^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = I
      | r = \frac {E_0 \paren {R \sin \omega t - L \omega \cos \omega t} } {R^2 + \omega^2 L^2} + \paren {I_0 + \frac {E_0 L \omega} {R^2 + \omega^2 L^2} } e^{-R t / L}
      | c = 
}}
{{end-eqn}}
The result follows by application of Multiple of Sine plus Multiple of Cosine: Sine Form.
{{qed}}
Theorem: Current in Electric Circuit/L, R in Series/Condition for Ohm's Law

Ohm's Law is satisfied by $K$ whenever the current $I$ is at a maximum or a minimum.

Proof:
From Electric Current in Electric Circuit: L, R in Series:
:$L \dfrac {\d I} {\d t} + R I = E$
defines the behaviour of $I$.
Let $I$ be at a maximum or a minimum.
Then from the Interior Extremum Theorem:
:$\dfrac {\d I} {\d t} = 0$
and so:
{{begin-eqn}}
{{eqn | l = E
      | r = 0 + R I
      | c = 
}}
{{eqn | r = R I
      | c = 
}}
{{end-eqn}}
which is Ohm's Law.
{{qed}}
Theorem: Current in Electric Circuit/L, R in Series/Minimum Current implies Increasing EMF

Let the current $I$ be at a minimum.
Then the EMF $E$ is increasing.

Proof:
From Electric Current in Electric Circuit: L, R in Series:
:$L \dfrac {\d I} {\d t} + R I = E$
defines the behaviour of $I$.
Taking the derivative:
:$L \dfrac {\d^2 I} {\d t^2} + R \dfrac {\d I} {\d t} = \dfrac {\d E} {\d t}$
From Second Derivative of Real Function at Minimum:
:$L \dfrac {\d^2 I} {\d t^2} \ge 0$
while from the Interior Extremum Theorem:
:$R \dfrac {\d I} {\d t} = 0$
and so:
:$\dfrac {\d E} {\d t} \ge 0$
The result follows from Increasing Function has Positive Derivative.
{{qed}}
Theorem: Current in Electric Circuit/L, R in Series/Maximum Current implies Decreasing EMF

Let the current $I$ be at a maximum.
Then the EMF $E$ is decreasing.

Proof:
From Electric Current in Electric Circuit: L, R in Series:
:$L \dfrac {\d I} {\d t} + R I = E$
defines the behaviour of $I$.
Taking the derivative:
:$L \dfrac {\d^2 I} {\d t^2} + R \dfrac {\d I} {\d t} = \dfrac {\d E} {\d t}$
From Second Derivative of Real Function at Minimum:
:$L \dfrac {\d^2 I} {\d t^2} \le 0$
while from the Interior Extremum Theorem:
:$R \dfrac {\d I} {\d t} = 0$
and so:
:$\dfrac {\d E} {\d t} \le 0$
The result follows from Decreasing Function has Negative Derivative.
{{qed}}
Theorem: Second Order ODE/y y'' = (y')^2

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

Proof:
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as:
{{begin-eqn}}
{{eqn | l = y p \frac {\d p} {\d y}
      | r = p^2
      | c = where $p = \dfrac {\d y} {\d x}$
}}
{{eqn | ll= \leadsto
      | l = y \frac {\d p} {\d y}
      | r = p
      | c = 
}}
{{eqn | ll= \leadsto
      | l = p = \dfrac {\d y} {\d x}
      | r = C_1 y
      | c = First Order ODE: $x \rd y = k y \rd x$
}}
{{eqn | ll= \leadsto
      | l = y
      | r = C_2 e^{C_1 x}
      | c = First Order ODE: $\d y = k y \rd x$
}}
{{end-eqn}}
{{qed}}
Theorem: First Order ODE/(1 - x y) y' = y^2

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:
{{begin-eqn}}
{{eqn | l = \frac {\d x} {\d y}
      | r = \frac {1 - x y} {y^2}
      | c = 
}}
{{eqn | n = 2
      | ll= \leadsto
      | l = \frac {\d x} {\d y} + \frac x y
      | r = \frac 1 {y^2}
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \int \map P y \rd y
      | r = \int \dfrac 1 y \rd y
      | c = 
}}
{{eqn | r = \ln y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = e^{\int P \rd y}
      | r = y
      | c = 
}}
{{end-eqn}}
Thus from Solution by Integrating Factor, $(2)$ can be rewritten as:
{{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d y} } {x y}
      | r = \dfrac 1 y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x y
      | r = \int \frac {\d y} y
      | c = 
}}
{{eqn | r = \ln y + C
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: First Order ODE/(2 x + 3 y + 1) dx + (2 y - 3 x + 5) dy = 0

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:
{{begin-eqn}}
{{eqn | l = \frac {\d w} {\d z}
      | r = \frac {2 z + 3 w} {3 z - 2 w}
      | c = 
}}
{{eqn | r = \frac {2 + 3 w / z} {3 - 2 w / z}
      | c = dividing top and bottom by $x$
}}
{{eqn | r = \frac {2 + 3 u} {3 - 2 u}
      | c = substituting $u$ for $w / z$
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \ln z
      | r = \int \frac {\d u} {\dfrac {2 + 3 u} {3 - 2 u} - u} + C
      | c = 
}}
{{eqn | r = \int \frac {\paren {3 - 2 u} \rd z} {2 + 3 u - u \paren {3 - 2 u} } + C
      | c = 
}}
{{eqn | r = \int \frac {\paren {3 - 2 u} \rd v} {2 \paren {1 + u^2} } + C
      | c = 
}}
{{eqn | r = \dfrac 3 2 \int \dfrac {\d u} {1 + u^2} - \int \dfrac {u \rd u} {1 + u^2} + C
      | c = 
}}
{{eqn | r = \dfrac 3 2 \arctan u - \dfrac 1 2 \, \map \ln {1 + u^2} + C
      | c = 
}}
{{end-eqn}}
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:
:$3 \, \map \arctan {\dfrac {y + 1} {x - 1} } = \map \ln {\paren {y + 1}^2 + \paren {x - 1}^2} + C$
{{qed}}