# 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 \( u(x) \) |
---|---|

Constant | Constant |

Polynomial of \( n \) degree | Polynomial 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

$$ f_{yy} > 0 $$

#### Saddle Point

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:

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

### Quadratic Factors

General form:

$$ \frac{As+B}{quadratic} + \frac{C}{linear} $$

## Using Laplace Transforms to Solve ODEs

For an ODE

$$ \frac{dy}{dx} = f(x) $$

Take the Laplace transform of the entire ODE

$$ \mathcal{L} \bigg\{ \frac{dy}{dx} \bigg\} = \mathcal{L} \{ f(x) \} $$

Rearrange and solve for \( \bar{y} \), invert Laplace transform back to answer.