From a differential equation to a function

Ordinary differential equations: Introduction

From a differential equation to a function

One can compute a derivative of a 'neat' function. Consider, for example, the exponential function \[y(t)=e^t\] whose derivative is again the exponential function \[y'(t)=e^t\tiny.\] The exponential function \(y(t)=e^t\) satisfies therefore at any time \(t\) the differential equation \(y'=y\). This is called a solution of the ODE.

The equation \(y'=y\) has other solutions, for example: \[\begin{aligned} y(t)&= 2e^t,\\ y(t)&= -e^t,\\ y(t)&= -\tfrac{1}{3}e^t. \end{aligned}\] All solutions have the same shape \[y(t)=c\cdot e^t\] for some constant \(c\). Are these all solutions of the differential equation? The answer is yes.

The general solution of the differential equation \[\frac{\dd y}{\dd t}=y\] is \[y(t)=c\cdot e^t\] for some constant \(c\).

Suppose that there are still solutions of the form \[y(t)=z(t)\cdot e^t\] where \(z(t)\) is an unknown function in \(t\).
On the one hand (using the product rule for differentiation) we have: \[y'(t)=\left(z(t)\cdot e^t\right)'= z'(t)\cdot e^t + z(t)\cdot \bigl(e^t\bigr)'= z'(t)\cdot e^t + z(t)\cdot e^t\] On the other hand, the function \(y(t)\) must satisfy the given differential equation \(y'=y\) and therefore \[y'(t)=z(t)\cdot e^t\tiny.\] From the last two formulas for \(y'(t)\) follows that \[z'(t)\cdot e^t =0\] must hold for all \(t\). This is only possible if \[z'(t)=0\] In other words, the function \(z(t)\) is constant.

All solutions of the differential equation \(y'=y\) have been found in explicit form.

In general, although a differential equation can have multiple solutions, it is sometimes still possible to write down the general solution of a differential equation in mathematical terms by using constants. Such constants are frequently called integration constants, simply because solving differential equations often boils down to integration of functions (for example, in solving by separation of variables).