Solving differential equations



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Linear ODEs with constant coefficients

The first method of integrating linear ordinary differential equations with constant coefficients is due to Euler, who made the solution of the form

$\frac {d^{n}y} {dx^{n}} + A_{1}\frac {d^{n-1}y} {dx^{n-1}} + \cdots + A_{n}y = 0$

depend on that of the algebraic equation of the nth degree,

$F(z) = z^{n} + A_{1}z^{n-1} + \cdots + A_n = 0$

in which z^k takes the place of

$\frac {d^{k}y} {dx^{k}}\quad\quad(k = 1, 2, \cdots, n).$

This equation F(z) = 0, is the "characteristic" equation considered later by Monge and Cauchy.

Example

y'''' - 2 y''' + 2 y'' - 2 y' + y = 0

has the characteristic equation

$z 4 - 2 z 3 + 2 z 2 - 2 z + 1 = 0$ .

This has zeroes, i, −i, and 1 (multiplicity 2). The solution basis is then

$e i x$ , $e - i x$ , $e x$ , $x e x$ .

This corresponds to the real-valued solution basis

$cos x$ , $sin x$ , $e x$ , $x e x$ .

If z is a (possibly complex) zero of F(z) of multiplicity m and $k\in\{0,1,\dots,m-1\}$ then $y = x k e z x$ is a solution of the ODE.

If the A_i are real then real-valued solutions are preferable. Since the complex z values will come in conjugate pairs, so will their corresponding y values; replace each pair with their linear combinations $\Re y$ and $\Im y$ .

A case that involves complex ( $\mathbb{C}$ ) root can be solved with the aid of Euler's formulae. Recall that Maclaurin series are defined as:

$e^x = \sum_{k = 0}^\infty {\frac{{x^{k} }}{{k!}}}$ ,

$\cos x = \sum_{k = 0}^\infty {\frac{{\left( { - 1} \right)^k x^{2k} }}{{\left( {2k} \right)!}}}$ , $\sin x = \sum_{k = 0}^\infty {\frac{{\left( { - 1} \right)^k }}{{\left( {2k + 1} \right)!}}x^{2k + 1} }$

And since

$\begin{matrix} i = \sqrt { - 1} \\ i^2 = - 1 \\ i^3 = - i \\ i^4 = 1 \\ \end{matrix} , e^{i\theta } = \sum_{k = 0}^\infty {\frac{{\left( i \right)^k }}{{k!}}\theta ^k = } \sum_{k = 0}^\infty {\frac{{\left( { - 1} \right)^k }}{{\left( {2k} \right)!}}\theta ^{2k} + i} \sum_{k = 0}^\infty {\frac{{\left( { - 1} \right)^k }}{{\left( {2k + 1} \right)!}}\theta ^{2k + 1} = } \cos \theta + i\sin \theta$

Giving the Euler's Formulae, $e i θ = cosθ + i sinθ$

Example: Suppose $P (D) y = 0$ for P(D)= $D 2 - 4 D + 5$

(Note: Here operator's notation is used to represent the linear ODE, y"-4y'+5=0),

Complete the square to find $\mathbb{C}$ roots by writing above Eq. in form:

$P(D)=\left[ {P - a} \right] + b^2$ roots are $r = a \pm bi.$

$P(D) = \left[ {D^2 - 4D + 4} \right] + 1 = \left[ {D - 2} \right]^2 + 1^2.\ \mathrm{Here}\ r = 2 \pm i$

are the characteristic roots. Hence solution in the form of $y = e r x$ are to be written as

$e^{\left( {2 + i} \right)x} = e^{2x + ix} = e^{2x} e^{ix} = e^{2x} \left( {\cos x + i\sin x} \right) = e^{2x} \cos x + ie^{2x} \sin x$

We think of $r = 2 \pm i{\rm{ }}$ as a root of multiplicity of 2. So seek two linearly independent solution to above equation yields:

$\left\{ {\begin{matrix} {y_1 = e^{2x} \cos x} \\ {y_2 = e^{2x} \sin x} \\\end{matrix}} \right.$

Any other solution to eq. has form of: $y c = c 1 e 2 x cos x + c 2 e 2 x sin x$ . Note the arbitrariness of C1 and C2 absorbs $\pm$ i.

Also, for repeated complex roots, multiply $y 1$ and $y 2$ repeatedly by x to generate a family of solutions, but only to multiplicity.

Solving linear equations