Second order linear ODE

Second order linear ODEs are in the following form:

$y''+p(t)y'+q(t)y=g(t)$

Important types of second order linear ODEs include

• Homogeneous
• Constant coefficients (where p and q are constants)

Initial value problem

There are two arbitrary constants in the solution of a second order linear ODE, so we need two initial conditions.

${\begin{cases}y(t_{0})=y_{0}\\y'(t_{0})=y_{0}'\end{cases}}$

Solutions

Constant coefficient, homogeneous

Whatever derived usually also works for variable coefficients.

These are the simplest kind. They have the general form

$ay''+by'+cy=0$

We can guess the form of the solution to be exponential, since exponential functions has the property of being the same after many differentiation.

$y=e^{rt}$

We substitute in the guess and obtain the characteristic equation

${\begin{aligned}ar^{2}e^{rt}+bre^{rt}+ce^{rt}&=0\\ar^{2}+br+c&=0\end{aligned}}$

Depending on the constants, it will give us anywhere from zero to two solutions: $y_{1}$ and $y_{2}$

There are three cases: 2 real roots, complex conjugates, and repeated roots, depending on the discriminant.

For 2 real roots, the two solutions are always independent so we get general solution easy.

For complex conjugates, the following real solutions can be derived from Euler's formula. Since complex number + complex conjugate becomes real.

${\frac {1}{2}}(y_{1}+y_{2})=e^{\lambda t}\cos(\mu t)$

${\frac {1}{2i}}(y_{1}-y_{2})=e^{\lambda t}\sin(\mu t)$

This gives us a set of real fundamental solutions.

For repeated roots, use reduction of order to find a second solution. The second solution is of the form

$te^{rt}$

Fundamental set of solutions

Given that two linearly independent solutions are given, the general solution is of the following form

$y(t)=c_{1}y_{1}+c_{2}y_{2}$

The independence of the solutions can be checked using the Wronskian. The two solutions are called a fundamental set of solutions.

This fundamental set always exists according to the General existence theorem.