Checking to see if a given function satisfies a given differential equation isn't too horrible of a task (at least for the functions you'll encounter in Calculus).
The task of solving differential equations from scratch is a bit different.
Differential equations fall into several very, very broad categories:
- ones you can solve right now by thinking backwards
- ones you'll be able to solve by the end of the calculus class
- ones you'll be able to solve if you take other courses about differential equations, and
- ones that people at places like MIT are still working on
We need to ignore most of these for now and concentrate on the ones you can solve by thinking backwards.
Practice:
Find a solution to the differential equation y' = 2x. | |
The expression 2x looks like it came from using the power rule get 2x after taking the derivative, we could have started with y = x2. This is, of course, only one possible solution. There are infinitely many other solutions. | |
Find all solutions to the differential equation y' = 2x. | |
Before taking the derivative, we could have started with y = x2. We could also have started with y = x2 + 1 or y = x2 - 9 or y = x2 + π. No matter what constant we stick on after x2, we'll still have a solution to the differential equation. This means we can write y = x2 + C, where C can be any constant, to take care of all solutions at once! At some point in the future, the constant + C in the example above will be known as the constant of integration. | |
Find a solution to the differential equation 
| |
Any solution to this d.e. must be its own derivative. The most obvious such function is y = ex. But we could also have y = 5ex, since then y' = 5ex also. Or we could have y = -4ex. Any equation of the form y = Cex (where C is a constant) is a solution to this differential equation. In the example above, the constant C can be anything, including zero. If C = 0 then y is a constant function; y = 0ex = 0 Since the derivative of the constant function y = 0 is also 0,
the differential equation 
is satisfied. | |
Find a solution to the d.e. y" = 5. | |
Here we need to think backwards twice. What was y'? We took the derivative of y' and got 5, so we could have started with y' = 5x If the derivative of y is 5x, then we could have started with 
| |
Find all solutions to the d.e. y" = 5. | |
Thinking backwards the first time, we have y' = 5x + B where B is any constant. Thinking backwards again, what function has a derivative of 5x + B? One answer is 
However, we could also add on any constant to this expression, since that wouldn't change the function's derivative. So the solution can be any function of the form 
| |
For the differential equation, find (a) one solution and then (b) all solutions.
y ' = 4x3
Answer
This looks like it came from using the power rule on the function
y = x3.
This is a solution. Since the derivative of a constant is zero, adding a constant to the function x3 doesn't change the derivative of that function. So we can write all solutions by
y = x3 + C
where C is any constant.
For the differential equation, find (a) one solution and then (b) all solutions.
y' = x5
Answer
When we take the derivative of a polynomial the degree drops by 1, so this almost looks like it came from taking the derivative of the function y = x6. However, the derivative of x6 is 6x5. To cancel out the extra 6 we get when taking the derivative, we need to introduce a factor of 
. A solution tothe differential equation could look like this:
All solutions look like this (for any constant C):
For the differential equation, find (a) one solution and then (b) all solutions.
Answer
What is a function whose derivative is sin x? One answer is
y = -cos x
(take the derivative of -cos x if you need to convince yourself). To write all solutions, we write
y = -cos x + C
where C is any constant.
For the differential equation, find (a) one solution and then (b) all solutions.
xy' = 4x
Answer
Divide both sides by x to get
y' = 4.
A solution is
y = 4x,
all solutions can be written as
y = 4x + C
where C is any constant.
For the differential equation, find (a) one solution and then (b) all solutions.
y" = -x-2
Answer
We need to think backwards twice. The first time, if we take the derivative of y' to get -x-2 we could have started with
Thinking backwards again, what function has 
as its derivative? This one:
y = ln x.
This is one solution. To find all solutions, start over. What has -x-2 as its derivative? Any function of the form
where B is any constant. In order to have such a y', we must have started with
y = ln x + Bx + C
where B,C are constants.
