SimpleOrdinaryDE.md

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• link: https://www.youtube.com/watch?v=BjvkBLfvkqY&list=PLMrJAkhIeNNTYaOnVI3QpH7jgULnAmvPA&index=4

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Simple Ordinary Differential Equation

First Order Linear Ordinary D.E

Let's say that bunnies are procreating
let x = bunny population
and the population grows at a rate $\lambda$

$\dot x = \frac{dx}{dt}=\lambda x, x(0) \ initial \ pop \ size$

We ask the question what is the population as a function of time ?
x(t) ?

Method 1:

$\frac{dx}{dt} = \lambda x$

$\frac{dx}{x} = \lambda dt$

$\int \frac{dx}{x} = \int \lambda dt$

=> $ln(x(t)) = \lambda t + C$
=> $x(t) = e^{\lambda t + C}$ we know that $e^{a+b}=e^{a}e^{b}$
=> $x(t) = e^{\lambda t}+K$
K ? $X(0) = e^0K \Rightarrow K = x(0)$

Second Order ODE

Example

The differential equation you've written down is a second-order linear differential equation with constant coefficients, which can be written in the standard form $\frac{d^2x}{dt^2} + \frac{k}{m} \frac{dx}{dt} + \frac{g}{m} x = 0$. To solve this type of differential equation, you can use the characteristic equation, which is given by the equation $\lambda^2 + \frac{k}{m} \lambda + \frac{g}{m} = 0$. The solutions to this equation are the so-called characteristic roots, which are the values of $\lambda$ that satisfy the equation. In this case, the characteristic roots are given by the quadratic formula:

$$\lambda = \frac{-\frac{k}{m} \pm \sqrt{\left(\frac{k}{m}\right)^2 - 4 \cdot \frac{g}{m}}}{2}.$$

Once you have the characteristic roots, you can find the general solution to the differential equation by writing it in the form $x(t) = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t}$, where $c_1$ and $c_2$ are constants and $\lambda_1$ and $\lambda_2$ are the characteristic roots. To find the specific solution to the equation, you need to use initial conditions, which specify the values of $x$ and $\dot{x}$ at a particular time $t_0$. Using these initial conditions, you can solve for the constants $c_1$ and $c_2$ and obtain the specific solution to the differential equation.

Maths