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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.
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.
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:
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
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.
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.)
Solve the differential equation dy/dx = 2xy
if (0, 5) lies on the graph.
ln y = x2 + c , (y > 0)
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.)
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