There is no general method for solving all differential equations. Instead, just as with integration, there is a collection of techniques, each of which applies to a particular class of equations. We'll look at a couple of these techniques. The first one applies to first order linear equations. The general first order linear equation can be written as y' + p(t) y = q(t) where p(t) and q(t) are arbitrary functions of t.
The solution is as follows: let This is called an integrating factor. Then the solution is
Here C is the arbitrary constant that goes along with the indefinite integral. You might ask, why isn't there another arbitrary constant with the integral in ? There could be one, but it wouldn't matter: different choices of that constant would end up giving you the same solutions. But the constant C is important, because different choices of C give different solutions.
Example 1: Solve
Here p(t) = 2 t and q(t) = 4 t. First step: find the integrating factor.
Second integration:
So the answer is:
If you are given an initial value y(t0) = y0 in addition to the differential equation, you have an initial value problem. Then after obtaining the general solution as above, you use the initial value to determine what C is. For example, in the last equation, suppose we had the initial value y(1)=4. Then we would set y=4, t=1 to get so . Thus the solution is
Why it works
For convenience I'm going to leave out all the (t)'s, so just remember that everything is supposed to be a function of t. Suppose we multiply the equation by :
Note that .We cleverly defined so that . So the left side of the equation becomes .Now we can integrate both sides:
Divide by , and you have the solution formula.
Special Cases:
The formula becomes a bit simpler if q(t) = 0: the differential equation is then called a homogeneous linear equation. Of course the antiderivative of is 0 + C (don't forget that C). So the solution is
For example:
We then have
so the solution is
Another important special case is when p is a constant. Then we have a constant-coefficient linear equation. The first integration is easy:
For example, we solve the initial value problem:
The integrating factor is , so we have to integrate
(how did I get that?)Thus the general solution is
Plugging in the initial conditions: 1 = 0 - 1/9 + C, C = 10/9 so the answer is