# MM217

### Log Rules

$$e^{\ln{(A)}} = A$$
$$B \ln{(A)} = \ln{(A^B)}$$
$$\ln{(AB)} = \ln{(A)} + \ln{(B)}$$
$$\ln{( \frac{A}{B})} = \ln{(A)} - \ln{(B)}$$

### Indices Rules

$$A^m \times A^n = A^{m+n}$$
$$\frac{A^m}{A^n} = A^{m-n}$$
$$(A^m)^n = A^{mn}$$
$$A^0 = 1$$
$$A^{-m} = \frac{1}{A^m}$$
$$A^{\frac{m}{n}} = \sqrt[n]{A^m}$$

## First Order Differential Equations

### First Order Separable

General Form:
$$f(y) \frac{dy}{dx} = g(x)$$

Integrate both sides wrt $$dx$$
$$\int f(y) \frac{dy}{dx} dx = \int g(x) dx$$

Simplify and evaluate
$$\int f(y) dy = \int g(x) dx$$
$$y = h(x) + C$$

### First Order Linear

General Form:
$$\frac{dy}{dx} + p(x)y = q(x)$$

The integrating factor (IF), $$R$$, is given by
$$R = e^{\int p(x) dx}$$

Multiply into general equation
$$R \frac{dy}{dx} + Rp(x)y = Rq(x)$$

Rewrite as
$$\frac{d}{dx} (yR) = Rq(x)$$

Simplify and evaluate
$$yR = \int R q(x) dx$$
$$y = h(x) + C$$

### First Order Homogeneous

General Form:
$$\frac{dy}{dx} = f(\frac{y}{x})$$

Let $$v = \frac{y}{x}$$, so that $$y = xv$$
$$\frac{dy}{dx} = \frac{d}{dx} (xv)$$

Since $$v$$ is a function of $$x$$, the differentiation product rule gives
$$\frac{d}{dx} (xv) = v + x \frac{dv}{dx}$$

Substituting back into the general equation
$$v + x \frac{dv}{dx} = f(v)$$

Solve as a first order separable equation
$$x \frac{dv}{dx} = f(v) - v$$

## Second Order Differential Equations

General Form:

$$a \frac{d^2 y}{dx^2} + b \frac{dy}{dx} + cy = f(x)$$

### Homogeneous

Where $$f(x) = 0$$ and the auxiliary equation is defined by
$$am^2 + bm + c = 0$$

The roots of which can be found by either factorisation or the quadratic formula
$$m = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$$

If the roots are:

#### Real and Distinct

$$(m_1, m_2) \in \mathbb{R}; m_1 \neq m_2$$

The General Solution will be of the form
$$y = A e^{m_1 x} + B e^{m_2 x}$$

#### Real and Coincident

$$(m_1, m_2) \in \mathbb{R}; m_1 = m_2$$

The General Solution will be of the form
$$y = (Ax + B) e^{mx}$$

#### Complex Conjugates

$$(m_1, m_2) \in \mathbb{C} ; m = p \pm iq$$

The General Solution will be of the form
$$y = e^{px} (A \cos (qx) + B \sin (qx))$$

### Inhomogeneous

Where $$f(x) \neq 0$$

The Complementary Function (CF), $$Y(x)$$, is defined as the General Solution of the corresponding homogeneous equation.

The form of the Particular Integral, $$u(x)$$, can be determined depending on the form of $$f(x)$$.

The General Solution of the inhomogeneous equation is then
$$y = Y(x) + u(x)$$

For $$f(x)$$ Try $$u(x)$$
ConstantConstant
Polynomial of $$n$$ degreePolynomial of $$n$$ degree
$$\sin (ax)$$ $$\alpha \sin (ax) + \beta \sin (ax)$$
$$a e^{bx}$$ $$\alpha e^{bx}$$
$$e^x \sin (2x)$$ $$Re ( \alpha e^{(p+iq)x} )$$

If $$f(x)$$ involves a term that is also in the Complementary Function, try $$PI = u = x \times (normal form of u)$$

#### Testing the Particular Integral

For a testing PI, $$u$$, find $$u'$$ and $$u''$$
Substitute into general equation for
$$\frac{d^2 y}{dx^2} = u''; \frac{dy}{dx} = u'; y = u$$
$$au'' + bu' + cu = f(x)$$

## Partial Differentiation

For a function $$f(x,y,z)$$, the partial derivative can be taken such that $$\frac{\partial f}{\partial x}$$ is the derivative of $$f(x,y,z)$$ with respects to $$x$$ where $$(y,z)$$ are constants.

Higher order partial differentiation notation is as follows

$$\frac{\partial ^3 f}{\partial x^3} = f_{xxx}$$

$$\frac{\partial ^2 f}{\partial x \partial y} = \frac{\partial}{\partial x} \bigg(\frac{\partial f}{\partial y} \bigg) = f_{yx}$$

$$\frac{\partial ^4 f}{\partial d \partial c \partial b \partial a} = \frac{\partial}{\partial d} \bigg( \frac{\partial}{\partial c} \bigg( \frac{\partial}{\partial b} \bigg( \frac{\partial f}{\partial a} \bigg) \bigg) \bigg) = f_{abcd}$$

### Approximations / Small Changes

For a function $$z = f(x,y)$$. If $$x$$ changes to $$x + \Delta x$$ and $$y$$ changes to $$y + \Delta y$$

$$\Delta z \approx \frac{\partial z}{\partial x} \Delta x + \frac{\partial z}{\partial y} \Delta y$$

Percentage change can be found by dividing each side by $$z$$

$$\frac{\Delta z}{z} \approx \frac{1}{z} \bigg[\frac{\partial z}{\partial x} \Delta x + \frac{\partial z}{\partial y} \Delta y \bigg]$$

### Extended Chain Rule

For a function $$z = (x,y)$$, where $$x = x(s,t)$$ and $$y = y(s,t)$$. A tree map of each dependency and the corresponding partial derivative can be drawn.

The higher partial derivative, $$\frac{\partial z}{\partial s}$$ can then be found by the sum of multiplying the partial derivatives of each branch containing the variable $$s$$.

$$\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s}$$

### Implicit Functions

For a function $$f(x,y,z)$$, where $$z = z(x,y)$$

$$\frac{\partial f}{\partial x} = z \frac{\partial z}{\partial x}$$

eg. For $$f(x,y,z) = x^2 + y^2 + z^2 = 1$$, where $$y$$ is an independent variable and $$z = z(x,y)$$

$$\frac{\partial f}{\partial x} = 2x + 2z \frac{\partial z}{\partial x}$$

### Directional Derivatives

For a function $$T = T(x,y,z)$$, at point $$P_0 (x_0, y_0, z_0 )$$ in the direction $$u$$.

The gradient of $$T$$ is given by:
$$\nabla T = \frac{\partial T}{\partial x} i + \frac{\partial T}{\partial y} j + \frac{\partial T}{\partial z} k$$

Evaluate $$\nabla T$$ at $$P_0$$.
Find the unit vector $$u$$
$$u = \frac{d}{|d|}$$

Where $$d$$ is the displacement, $$|d|$$ is the magnitude calculated by

$$|d| = \sqrt{i^2 + j^2 + k^2}$$

The directional derivative of $$T$$ at point $$P_0$$ in the direction $$u$$ can then be found by

$$\frac{dT}{ds} = \nabla T \cdot u$$

### Stationary Points

#### Local Maximum

If
$$f_{xx}f_{yy} - f_{xy}^2 > 0$$

and
$$f_{xx} < 0$$

or $$f_{yy} < 0$$

#### Local Minimum

If
$$f_{xx}f_{yy} - f_{xy}^2 > 0$$

and
$$f_{xx} > 0$$

or
$$f_{yy} > 0$$

If
$$f_{xx}f_{yy} - f_{xy}^2 < 0$$

## Double Integration

$$\int _{a}^{b} \int _{m}^{n} f(x,y)dy dx$$

Describes an area restricted by the lines $$y=m,y=n$$ and $$x=a,x=b$$.

### Reversing Integration Order

Draw the region $$R$$ and state the new boundary lines in the reverse order.

## Polar Coordinates

$$dxdy \rightarrow rdrd \theta$$

## Laplace Transforms

### Basic Properties

#### Linearity

$$\mathcal{L} \{f(t) + g(t) \} = \mathcal{L} \{ f(t) \} + \mathcal{L} \{ g(t) \}$$
$$\mathcal{L} \{a f(t) \} = a \mathcal{L} \{f(t)\}$$

#### First Shifting Property

$$\mathcal{L}\{ e^{-at} f(t) \} = \bar{f}(s+a)$$

#### Multiplication by t

$$\mathcal{L} \{tf(t) \} = - \frac{d}{ds} \big[ \mathcal{L} \{ f(t) \} \big]$$

### Inversion

If $$\bar{f} (s) = \mathcal{L} \{f(t) \}$$, then $$f(t) = \mathcal{L}^{-1} \{ \bar{f} (s) \}$$ and $$\mathcal{L} ^{-1}$$ is called the inverse Laplace transform of $$\bar{f} (s)$$.

## Partial Fractions

### Linear Factors

Factorise denominator into linear factors and solve for unknown numerators.

General form:
$$\frac{A}{linear} + \frac{B}{linear}$$

### Repeated Linear Factors

General form:
$$\frac{A}{linear} + \frac{B}{repeated} + \frac{B}{repeated^2}$$

$$\frac{As+B}{quadratic} + \frac{C}{linear}$$
$$\frac{dy}{dx} = f(x)$$
$$\mathcal{L} \bigg\{ \frac{dy}{dx} \bigg\} = \mathcal{L} \{ f(x) \}$$
Rearrange and solve for $$\bar{y}$$, invert Laplace transform back to answer.