
Differential equations
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Differential equations
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:
- Collect x's on one side; y's and dy/dx on the other.
- Integrate both sides with respect to x.
- 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 ex 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 = ex2 + c = ex2 . ec = Aex2
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 = 5ex2
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.)