Differential Equations

First Order

A first order differential equation has $\frac{d y}{d x}$ as it's highest derivative. For the two methods below, it is important the equation is in the correct form specified.

Seperating Variables

For an equation of the form $\frac{d y}{d x} = f (x) g (y)$

The solution is $\int \frac{d y}{g ( y )} = \int f (x) d x$

Integrating Factors

For an equation of the form $\frac{d y}{d x} + P (x) y = Q (x)$

An integrating factor $μ$ can be found such that: $μ (x) = e^{\int P (x) d x}$

Multiplying through by $μ$ gives

$μ (x) \frac{d y}{d x} + μ (x) P (x) y = μ (x) Q (x)$

Then, applying he product rule backwards gives a solution:

$\frac{d}{d x} (μ (x) y) = μ (x) Q (x)$ $μ (x) y = \int μ (x) Q (x) d x$

Second Order

A second order ODE has the form: $a \frac{d ^{2} y}{d x ^{2}} + b \frac{d y}{d x} + cy = f (x)$

The equation is homogeneous if $f (x) = 0$ .

The auxillary equation is $a k^{2} + bk + c = 0$

This gives two roots $k_{1}$ and $k_{2}$ , which determine the complementary function:

Roots	Complementary Function
$k_{1}$ and $k_{2}$ both real	$y = A e^{k_{1} x} + B e^{k_{2} x}$
$k_{1} = k_{2}$ , both real	$y = (A + B x) e^{k x}$
$k_{1} = α + β i$ and $k_{2} = α - β i$	$y = e^{αx} (A cos β x + B sin β x)$

The complementary function is the solution. Sometimes, initial conditions will be given which allow the constants $A$ and $B$ to be found.

If the system is non-homogenous, ie $f (x) \neq = 0$ , then a particular integral is needed too, and the solution will have form $y = c . f . + p . i .$ . The particular integral is found using a trial solution, then substituting it into the equation to find the coefficients. Note that if the particular integral takes the same form as the complementary function, an extra $x$ will need to be added to the particular integral for it to work, it $a e^{k x}$ would become $a x e^{k x}$

$f (x)$	Trial Solution
const $k$	const $α$
polynomial $a x^{r} + ... + b x + c$	$α x^{r} + ... + β x + γ$
$a cos k x$ or $a sin k x$	$α cos k x + β sin k x$
$a e^{k x}$	$α e^{k x}$

Example

$\frac{d ^{2} y}{d x ^{2}} + 4 y = sin x y (0) = 1, \frac{d y}{d x} (0) = 1$

Auxillary equation: $k^{2} + 4 = 0$ $k = \pm 2 i$

Complementary function is therefore: $y = A cos 2 x + B sin 2 x$

System is non-homogeneous, so have to find a particular integral. For this equation $f (x) = sin x$ , so the p.i. is $y = a cos x + b sin x$ . $y = a cos x + b sin x$ $\frac{d y}{d x} = - a sin x + b cos x$ $\frac{d ^{2} y}{d x ^{2}} = - a cos x - b sin x$

Substituting this into the original equation: $\frac{d ^{2} y}{d x ^{2}} + 4 y = - a cos x - b sin x + 4 a cos x + 4 b sin x = sin x$

Comparing coefficients: $cos x : 3 a = 0 \Rightarrow a = 0$ $sin x : 3 b = 1 \Rightarrow b = 1/3$

The general solution is therefore: $y = c . f . + p . i . = A cos 2 x + B sin 2 x + \frac{1}{3} sin x$

Using initial conditions to find constants, for $y (0) = 1$ $1 = A cos 0 + B sin 0 + \frac{1}{3} sin 0 \Rightarrow A = 1$

For $\frac{d y}{d x} (0) = 1$ $\frac{d y}{d x} = - 2 A sin 2 x + 2 B cos 2 x + \frac{1}{3} cos x$ $1 = - 2 A sin 0 + 2 B cos 0 + \frac{1}{3} cos 0$ $1 = 2 B + \frac{1}{3} \Rightarrow B = \frac{1}{3}$

Particular solution for given initial conditions is therefore: $y = cos 2 x + \frac{1}{3} sin 2 x + \frac{1}{3} s in x$

Computer Systems Engineering Notes