Introduction to Differential Equations

Part 2: Initial value problems

Our mathematical model describing the spread of the rumor consists of two parts. The first is the differential equation

This second condition is called an initial condition -- the value of dependent variable (in this case S) at the first value of the independent variable (in this case t) under consideration. A differential equation together with an initial condition is called an initial value problem. The implied "problem" is to find a function that satisfies both conditions.

We need to discuss the term "solution." A solution to a differential equation is a function that satisfies the relation for all values of the independent variable under consideration. For example, Y(t) = exp(t) is a solution of the differential equation

The function Y(t) = 3 exp(t) is another solution to this differential equation.

A solution to an initial value problem is a solution to the differential equation that also satisfies the initial condition. So, Y(t) = exp(t) is a solution to the initial value problem

but, Y(t) = 3 exp(t) is not.

One can find a symbolic description for the solution of the initial value problem

Here is such a description:

For now, we will not discuss how you could find such a solution. Rather, we will just verify that it works.

1. Use your computer algebra system to check that the given function S is indeed a solution to the initial value problem
   dS/dt = k S (M - S) for all nonnegative t and S(0) = 2.

   You may have to simplify your expressions for the two sides of the differential equation to show they are the same.

2. Explain why the following functions are not solutions to the given initial value problems.

   	Function	Initial Value Problem
   a)	Y(t) = 4 exp(t) - t - 1	dY/dt = t + Y,  Y(0) = -1
   b)	Y(t) = 1/(t - 1)	dY/dt = Y2,    Y(0) = -1

A differential equation usually has infinitely many solutions. This should not be surprising when we realize that finding the family of all antideriavtives for a function f is the same as finding all solutions Y to the differential equation dY/dt = f(t). Indeed, a procedure for finding all solutions of a first-order differential equation usually involves an antidifferentiation step and so the introduction of an arbitrary constant.

3. Find all solutions of the differential equation
   dY/dt = t² + 1.

4. Find the solution to the initial value problem
   dY/dt = t² + 1    with     Y(0) = 3.