# Differential equations

An ordinary differential equation, abbreviated ODE, is an equation that involves derivatives of a function. The order of the equation is the order of the highest derivative in the equation.

ODE are used frequently to model real-world phenomena.

- The simplest ODE defines the
**primitive**$$x$$ of a function $$f$$ $$$ \frac{d}{dt} x(t) = f(t)$$$ - The
**rate of decay**of radioactive atoms is proportional to their number $$N(t)$$ over time. It satisfies the**first order ODE**$$$\dot{N} = -\lambda N$$$ (the dot represents the derivative with respect to time). - The force on a string is inversely proportional to its position $$x(t)$$ away from its equilibrium. By
**Newton's law**, its acceleration $$\ddot{x}$$ is proportional to the force. This gives the**second order ODE**$$$m \ddot{x} = - k x.$$$ - The
**Volterra equation**models the evolution of a population of prey $$x(t)$$ and predators $$y(t)$$. It is a first order ODE. $$$ \dot{x} = x(1-y),\quad \dot{y} = -y(1-x).$$$

In general, a solution to a differential equation is not unique. For instance the solutions to the equation $$$\dot{x} = -x$$$ are the functions of the form $$$ x(t) = x_0 e^{-t}$$$ for any $$x_0$$. Indeed, $$$\dot{x}(t) = -x_0e^{-t} = -x(t).$$$

- A general solution to the ODE is the collection of all the solutions expressed using a parameter. Here $$x_0 e^{-t}$$ is the solution, and the parameter is $$x_0$$.
- A particular solution is a solution for a given value of the parameter, for example $$2e^{-t}$$.
- The solution of a first order ODE is unique if we impose an initial condition, i.e. the value of the solution at a given point. For instance the solution to the ODE with initial condition $$x(1)=1$$ is $$e^{1-t}$$.
- The number of parameters is generally equal to order of the ODE.

We detail classical techniques for solving differential equations.

- Direct integration The solutions to $$$\dot{x}(t) = f(t)$$$ are the primitives $$\displaystyle x(t) = \int f(t)\,dt$$.
The solutions to the $$\ddot{x}(t) = g$$ are obtained by integrating twice. This gives $$x(t) = g\, t^2/2 + x_1 t + x_0$$.

- Separation of variables The solutions to $$$\dot{x}(t) = \frac{g(t)}{f(x)}$$$ satisfy $$f(x)\,dx = g(t)\,dt$$. Thus, $$\displaystyle \int f(x)\,dx = \int g(t)\,dt. $$
The solutions to $$\dot{x} = 2xt $$ solve $$dx / x = 2t\,dt$$. We integrate both sides to find $$\ln\vert x\vert = t^2 + c$$, hence $$ x = c' e^{t^2}$$.

- Reduction by substitution where we change the function to be differentiated.
To solve $$\dot{x} - 2 t x = \cos t$$, we set $$y = e^{-t^2}x$$. We get $$$\dot{y} = e^{-t^2}\dot{x} -2t e^{-t^2}x = e^{-t^2}(\dot{x} - 2t x) = e^{-t^2}\cos t. $$$ So, $$\displaystyle y(t) = \int e^{-t^2}\cos t\,dt.$$