A differential equation is where we have to find the original equation of a function from an equation involving its gradient.

These are equations where dy/dx is in terms of x.

In these situations we just use regular integration to find the original equation.

Example:

Find the equation of the graph whose gradient,

Having done our integration we get an expression that includes the constant '+ c'.

If we know one point we can find c, and this will give us a particular solution.

If we do not know a point on the graph then we cannot find c and we get a general solution.

General Solutions

In the example above we found a general solution y = x² + 2x + c. To illustrate a general solution we draw a family of curves on a coordinate grid, each graph representing a different value for c. The family of curves for y = x² + 2x + c is:

Particular Solutions

A Particular Solution, (or Particular Integral), is a specific solution to the question that is found using an extra piece of information - one point that lies on the graph.

For instance, if the graph y = x² + 2x + c contains the point (-2, 3), then we know that when x = -2, y = 3.

This gives us:

4 − 4 + c = 3

Therefore:

c = 3 and the particular solution is y = x² + 2x + 3.

Sometimes we cannot use simple integration to find the general solution to a differential equation. This is because some differential equations are in terms of x and y! In order to solve these equations we:

1. Collect x's on one side; y's and dy/dx on the other.
2. Integrate both sides with respect to x.
3. Simplify the resulting equation.

Example 1

Solve the differential equation

1. Rearrange to get, ydy/dx = 3 x².

2. Integrate to get

This solves to give y² / 2 = x³ + c

3. Simply this to get y² = 2x³ + c

(If you are worried about the fact that we have not changed the constant after multiplying both sides by 2 you can re-christen it with another letter, say k. Often when working with e^x the constant can be included in the expression and is normally given the letter A.)

Example 2

Solve the differential equation dy/dx = 2xy

if (0, 5) lies on the graph.

ln y = x² + c , (y > 0)

Therefore:

y = e^{x2 + c} = e^x2 . e^c = Ae^x2

This is the general solution.

To find the particular solution we use the fact that when x = 0, y = 5. This gives us A = 5, and the particular solution: y = 5e^x2

Note: Notice that we had to include the restriction that y > 0 as ln y is only defined if y > 0. (This is because you cannot find the log of a negative number.)