Integrating a function

Ordinary differential equations: Separable differential equations

Integrating a function

Solving an ODE of the form dy/dt = f(t) A direct application of integration is the solution of an ordinary differential equation of type

$\frac{\dd y}{\dd t}=f(t)$ for some function $f(t)$ . The general solution of this first order ODE can be found by integration:

$y(t)=\int f(t)\,\dd t\tiny.$

A stronger suggestion of integration comes from rewriting the differential equation in terms of differentials:

$\dd y=f(t)\,\dd t\tiny.$ The variables $y$ and $t$ are now separated from each other: $\dd y$ is to the left and everything with $t$ in it to the right. Integrating the left- and right-hand sides yields:

$\int \dd y=\int f(t)\,\dd t\quad \text{i.e.}\quad y = \int f(t)\,\dd t\tiny.$ The integration constants separately do not provide new solutions and can therefore be combined into a single integration constant. This is in the final formula already done secretly.

Example The first-order ODE

$\frac{\dd y}{\dd t}=3t^2$ is in differential form

$\dd y=3t^2\,\dd t.$ The right-hand side of this equation can also be written as $\dd(t^3)$ , because $t^3$ is an antiderivative of $3t^2$ . So we have an equality between two differentials:

$\dd y=\dd(t^3).$ The functions ‘behind the d's’ are thus equal to each other up to a constant:

$y=t^3+c$ for some constant $c$ .

You may also apply integration to the left- and right-hand side in

$\dd y=3t^2\,\dd t$ Then you get

$\int \dd y=\int 3t^2\,\dd t$ and the left- and right-hand side are equal to $y$ and $t^3$ , respectively, up to an integration constant.

General solution method for an initial value problem When you have an initial value problem

$\frac{\dd y}{\dd t}=f(t),\quad y(t_0)=y_0$ for a certain function $f(t)$ and parameters $t_0$ and $y_0$ , then you can also work with definite integrals in order to determine the solution of the initial value problem:

$\int_{y_0}^{y(t)}d\eta=\int_{t_0}^t f(\tau)\,d\tau$ So:

$y(t) = y_0+ \int_{t_0}^t f(\tau)\,d\tau$