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
$$ f_{yy} > 0 $$

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.