Seperation of Variables {#header-n2889}
=======================

Differentiation Rules {#header-n2891}
---------------------

The Chain Rule and Product Rule can be established visually fairly
easily [^1] and the proofs are fairly straightforward. [^2]

### Product Rule {#header-n2896}

$$\begin{aligned}
\frac{\operatorname{d} }{\operatorname{d} x}\left( u\cdot v \right)&= \frac{\operatorname{d} u}{\operatorname{d} v}\cdot v + u \cdot \frac{\operatorname{d} v}{\operatorname{d} x} \label{rule_11}\\
\ \notag \\
\frac{\operatorname{d} }{\operatorname{d} x}\left( f\left( x \right)\cdot g\left( x \right) \right)&= f'\left( x \right)\cdot g\left( x \right)+ f\left( x \right)\cdot g'\left( x \right) \label{prodruledefleib}\end{aligned}$$

### Chain Rule {#header-n2898}

$$\begin{aligned}
\frac{\operatorname{d}y }{\operatorname{d} x} &= \frac{\operatorname{d} y}{\operatorname{d} u} \cdot \frac{\operatorname{d} u}{\operatorname{d} x}\\
\ \notag \\
\frac{\operatorname{d} }{\operatorname{d} x}\left[ f\left( g\left( x \right) \right) \right]&= f'\left( g\left( x \right) \right)\cdot g\left( x \right) \\
\ \notag \end{aligned}$$

Integration Rules {#header-n2901}
-----------------

### Integration by Substitution {#header-n2902}

The chain rule can be used for integration with some clever
substitution:

Let:

$$\begin{aligned}
\begin{matrix}
u &= g(x) & \quad F(x): \enspace F'(x) = f(x) = y \\
\frac{du}{dx} &= g'(x)
\end{matrix}\end{aligned}$$

Now by direct substitution into the chain rule:

$$\begin{aligned}
\frac{\operatorname{d} }{\operatorname{d} x}\left[ F'\left( u \right) \right]&= F'\left( g\left( x \right) \right)\cdot g'\left( x \right)\notag \\
&= f\left( g\left( x \right) \right)\cdot g'\left( x \right)\notag \\
\implies f\left( g\left( x \right) \right)\cdot g'\left( x \right)&= \frac{\operatorname{d} }{\operatorname{d} x}\left[ F\left( u \right) \right]\notag \\
f\left( g\left( x \right) \right)\cdot g'\left( x \right)&= \frac{\operatorname{d} }{\operatorname{d} x}\left[ F\left( u \right) +C \right]\notag \\ \end{aligned}$$

Now by integrating both sides:

$$\begin{aligned}
\int^{}_{} f\left( g\left( x \right) \right)\cdot g'\left( x \right) \operatorname{d}x &= \int^{}_{} \frac{\operatorname{d} }{\operatorname{d} x}\left[ F\left( u \right)+ C \right] \operatorname{d}x \notag \\
&= F\left(u \right) + C\notag \\
&= \int^{}_{} f\left( u \right) \operatorname{d}u\notag \end{aligned}$$

So what we have is integration by substitution:

$$\begin{aligned}
\int^{}_{} f\left( g\left( x \right) \right)\cdot g'\left( x \right) \operatorname{d}x &= \int^{}_{} f\left( u \right) \operatorname{d}u \label{ibysub1} \\
\int^{}_{} f\left( u \right)\cdot \frac{\operatorname{d}u }{\operatorname{d} x} \operatorname{d}x&= \int^{}_{} f\left( u \right) \operatorname{d}u \label{ibysubl} \end{aligned}$$

This basically means that if an integral looks like the differentials
could cancel out, they do, making the *Leibniz* notation particularly
useful.

### Integration by Parts {#header-n2914}

The product rule can be used for integration, but it's only fruitful
when:

1. You can choose some $u= f\left( x \right)$ that simplifies when
differentiated Or atleast stays the same e.g.
$\frac{\operatorname{d} }{\operatorname{d} x}\left[ \sin{\left( x \right)} \right]= \cos{\left( x \right)}$

2. $\operatorname{d} v= g'\left( x \right) \operatorname{d} x$ can be
chosen such that the differential can be readily integrated to give
$v$.

 \
Consider the Product Rule $\left( \ref{prodruledefleib} \right)$:

$$\begin{aligned}
\frac{\operatorname{d} }{\operatorname{d} x}\left[ f\left( x \right)\cdot g\left( x \right) \right]&= f'\left( x \right)\cdot g\left( x \right)+ f\left( x \right)\cdot g'\left( x \right)\notag \\
\end{aligned}$$

Let,

$$\begin{aligned}
\begin{matrix}
&u = f\left( x \right) &&v = g\left( x \right)\\
&\frac{\operatorname{d}u }{\operatorname{d} x}= f'\left( x \right) && \frac{\operatorname{d}v }{\operatorname{d} x} = g'\left( x \right)
\end{matrix}\end{aligned}$$

Now we have:

$$\begin{aligned}
\int^{}_{} \left( \frac{\operatorname{d}u }{\operatorname{d} x}\cdot v + u\cdot \frac{\operatorname{d}v }{\operatorname{d} x} \right) \operatorname{d}x &= u\cdot v\notag \\
\ \notag \\
\int^{}_{} \left( v\cdot \frac{\operatorname{d}u }{\operatorname{d} x} \right) \operatorname{d}x + \int^{}_{} \left( u\cdot \frac{\operatorname{d}v }{\operatorname{d} x} \right) \operatorname{d}x&= u\cdot v \notag \end{aligned}$$

By Rule $\left( \ref{ibysubl} \right)$ we have:

$$\begin{aligned}
\int^{}_{} v \operatorname{d}u + \int^{}_{} u \operatorname{d}v &= u\cdot v \notag \\
\int^{}_{} u \operatorname{d}v &= u\cdot v - \int^{}_{} v \operatorname{d}u \label{ibypartl} \end{aligned}$$

#### Application {#header-n2936}

These are really the only two rules we've got (other than manipulation
with partial fractions if possible) so the only trick is choosing when
to use which one:

- Look at the intergrand
$\int^{}_{} \left[\enspace \right] \operatorname{d}x$ :

- If it's of the
form$\left[ f\left( u \right)\cdot \frac{\operatorname{d}u }{\operatorname{d} x}\right] = \left[ f\left( g\left( x \right) \right)\cdot g'\left( x \right) \right]$

- Use Integration by Substitution

- If it's of the form:
$\left[ f\left( x \right)\cdot \frac{\operatorname{d}u }{\operatorname{d} x} \right]= \left[ f\left( x \right)\cdot g'\left( x \right) \right]$

- Use Integration by Parts

Introduction to Ordinary Differential Equations
-----------------------------------------------

Equations involving differentials like $\operatorname{d} y$ or
$\operatorname{d} x$ or $\frac{\operatorname{d}y }{\operatorname{d} x}$
are differential equations.

If the derivatives correspond to only a single independent variable and
do not involve partials e.g. ($\frac{\partial u }{\partial x}$) they are
said to be *Ordinary Differential Equations* (***ODE***).

#### Classifying Differential Equations

#### Order and Degree

- The Order of a differential corresponds to the higest derivative
taken

- The Degree is the highest power of the highest order derivative of
the *ODE*

#### Linearity

A Linear ODE is of the form:

$$\begin{aligned}
\sum^{n}_{0} \left[ a_0\left( x \right)\cdot \left( \frac{\operatorname{d}^ny }{\operatorname{d} x^n} \right) \right] \label{lindef}\end{aligned}$$

#### Solutions to Differential Equations

In order of preference, to solve a differential equaiton:

1. Solve for $y$ (explicit)

2. Solve for $x$ (explicit)

3. Solve for 0 (implicit)

Seperable Differential Equations
--------------------------------

A differential equation of the form: $$\begin{aligned}
g\left( y \right)\cdot \frac{\operatorname{d}y }{\operatorname{d} x} = f\left( x \right) \label{sepdiffform}\end{aligned}$$
Is a seperable Ordinary Differential Equation and has a solution:

$$\begin{aligned}
\int^{}_{} g\left( y \right) \operatorname{d}y = \int^{}_{} f\left( x \right) \operatorname{d}x \label{sepdiffsol} \end{aligned}$$

#### Proof

$$\begin{aligned}
g\left( y \right)\cdot \frac{\operatorname{d}y }{\operatorname{d} x} &= f\left( x \right)\notag \\
\implies \int^{}_{} g\left( y \right)\frac{\operatorname{d}y }{\operatorname{d} x} \operatorname{d}x &= \int^{}_{} f\left( x \right) \operatorname{d}x\notag \\ \end{aligned}$$

By the Substitution Rule at ($\ref{ibysubl}$): $$\begin{aligned}
\int^{}_{} g\left( y \right) \operatorname{d}y &= \int^{}_{} f\left( x \right) \operatorname{d}x \end{aligned}$$

Equations Reducible to Separable Equations
------------------------------------------

Some equations can be tricky to deal with, there is a method of
$u$-Substitution:

Take some equation of the form:
$$\frac{\operatorname{d}y }{\operatorname{d} x}= f\left( \frac{x}{y} \right)$$
We can perform the $u$-substitution:

$$\begin{aligned}
u&= \frac{y}{x}\notag \\
\implies y &= u\cdot x\notag \\
\implies \frac{\operatorname{d}y }{\operatorname{d} x}&= \frac{\operatorname{d}u }{\operatorname{d} x}\cdot x+ \left( 1 \right)\cdot u \notag\end{aligned}$$

Substituting in the terms: $$\begin{aligned}
\frac{\operatorname{d}y }{\operatorname{d} x}&= f\left( \frac{y}{x} \right)\notag \\
\frac{\operatorname{d}u }{\operatorname{d} x}\cdot x + u&= f\left( u \right)\notag \\
\frac{\operatorname{d}u }{\operatorname{d} x}\cdot x&= f\left( u \right)- u\notag \\
\frac{1}{f\left( u \right)- u}\cdot \frac{\operatorname{d}u }{\operatorname{d} x}\cdot x &= 1\notag \\
\frac{1}{f\left( u \right)- u }\cdot \frac{\operatorname{d}u }{\operatorname{d} x}&= \int^{}_{} \frac{1}{x} \operatorname{d}x \notag \\
\int^{}_{} \frac{1}{f\left( u \right)- u}\cdot \frac{\operatorname{d}u }{\operatorname{d} x} \operatorname{d}x&= \int^{}_{} \frac{1}{x} \operatorname{d}x\notag \\
\int^{}_{} \frac{1}{f\left( u \right)- u} \operatorname{d}u &= \ln{ \left| x \right| }+ c \end{aligned}$$

Now presume
$\exists G\left( u \right): G\left( u \right)= \int^{}_{} \frac{1}{f\left( u \right)- u} \operatorname{d}u$

$$\begin{aligned}
G\left( u \right)&= \ln{ \left| x \right| }+ c\notag \\
G\left( \frac{y}{x} \right)&= \ln{ \left| x \right| }+ c\notag \\
G\left( \frac{y}{x} \right)+ \ln{ \left| x \right| }+ c &= 0 \label{redsepfin}\end{aligned}$$

Hence by by ([\[redsepfin\]](#redsepfin){reference-type="ref"
reference="redsepfin"}) there must atleast be an implicit solution to
the equation assuming that the integral can be solved.

[^1]: [Visualizing the chain rule and product
rule](https://www.youtube.com/watch?v=YG15m2VwSjA)

[^2]: [Differentiation Rules
Proof](https://www.dropbox.com/s/by7pzzkap8gyvav/Differentiation%20Rules.pdf?dl=0)