Second order linear ODE: Difference between revisions

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Depending on the constants, it will give us anywhere from zero to two solutions: <math>y_1</math> and <math>y_2</math>
Depending on the constants, it will give us anywhere from zero to two solutions: <math>y_1</math> and <math>y_2</math>


Based on [[Abel's theorem]], the general solution is of the following form
Given that two linearly independent solutions are given, the general solution is of the following form


<math>
<math>
y(t) = c_1 y_1+c_2 y_2
y(t) = c_1 y_1+c_2 y_2
</math>
</math>
The independence of the solutions can be checked using the [[Wronskian]].

Revision as of 22:17, 21 May 2024

Second order linear ODEs are in the following form:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{cases} y(t_0)=y_0\\ y'(t_0)=y_0' \end{cases} }

Solutions

Constant coefficient, homogeneous

These are the simplest kind. They have the general form

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=e^{rt} }

We substitute in the guess and obtain the characteristic equation

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{aligned} ar^2e^{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: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_1} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_2}

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y(t) = c_1 y_1+c_2 y_2 }

The independence of the solutions can be checked using the Wronskian.