1. Solving Differential Equations
A differential equation (or "DE") contains derivatives or differentials.
Recall from the Differential section in the Integration chapter, that a differential can be thought of as a derivative where "dy/dx" is not written in fraction form.
Examples of Differentials
dx (this means "an infinitely small change in x")
dθ (this means "an infinitely small change in θ")
dt (this means "an infinitely small change in t")
Examples of Differential Equations
Example 1.
We saw the following example in the Introduction to this chapter. It involves a derivative (dy/dx):
As we did before, we would integrate it to produce the required solution. This will be a general solution (involving K, a constant of integration).
So we proceed as follows:
and this gives
But where did that dy go from the dy/dx? Why did it seem to disappear?
In this example, we appear to be integrating the x part only (on the right), but in fact we have integrated with respect to y as well (on the left). Differential equations are like that - you need to integrate with respect to two (sometimes more) different variables, one at a time.
We could have written our question only using differentials:
dy = (x2 − 3)dx
What we have done is to multiply both sides of the original question by dx.
Now we integrate both sides, the left side with respect to y (that's why we use "dy") and the right side with respect to x (that's why we use "dx") :
∫ dy = ∫(x2 − 3)dx
Then the answer is the same as before, but this time we have arrived at it considering the dy part more carefully:
On the left hand side, we have integrated ∫ dy = ∫ 1 dy to give us y.
Example 2
This example also involves differentials:
θ2 dθ = sin(t + 0.2) dt
We have:
A function of θ with dθ on the left side, and
A function of t with dt on the right side.
To solve this, we would integrate both sides, one at a time, as follows:
∫ θ2 dθ = ∫ sin(t + 0.2) dt
θ3/3 = −cos(t + 0.2) + K
We have integrated with respect to θ on the left and with respect to t on the right.
Solving a differential equation
From the above examples, we can see that solving a differential equation means finding an equation with no derivatives that satisfies the given differential equation. Solving a differential equation always involves one or more integration steps.
It is important to be able to identify the type of differential equation we are dealing with before we attempt to solve it.
Definitions
First order DE: Contains only first derivatives
Second order DE: Contains second derivatives (and possibly first derivatives also)
Degree: The highest power of the highest derivative which occurs in the differential equation.
Examples - Order and Degree
1) 
This DE has order 2 (the highest derivative appearing is the second derivative) and degree 1 (the power of the highest derivative is 1.)
2) 
This DE has order 1 (the highest derivative appearing is the first derivative) and degree 5 (the power of the highest derivative is 5.)
3) (y")4 + 2(y')7 − 5y = 3
This DE has order 2 (the highest derivative appearing is the second derivative) and degree 4 (the power of the highest derivative is 4.)
General and Particular Solutions
When we first performed integrations, we obtained a general solution (involving a constant, K).
We obtained a particular solution by substituting known values for x and y. These known conditions are called boundary conditions (or initial conditions).
It is the same concept when solving differential equations - find general solution first, then substitute given numbers to find particular solutions.
Let's see some examples of finding solutions of first order, first degree DEs.
Example 1
a. Find the general solution for the differential equation
dy + 7x dx = 0.
b. Find the particular solution given that y(0) = 3.
Example 2
Find the particular solution of
y' = 5
given that when x = 0, y = 2.
Example 3
Find the particular solution of
y''' = 0
given that:
y(0) = 3, y'(1) = 4, y''(2) = 6
Example 4
After solving the differential equation,
(we will see how to solve this DE in the next section Separation of Variables), we obtain the result
y = c ln x
Did we get the correct general solution?
Second Order DEs
We include two more examples here to give you an idea of second order DEs. We will see later in this chapter how to solve such Second Order Linear DEs.
Example 5
The general solution of the second order DE
y'' + a2y = 0
is
y = A cos ax + B sin bx
Example 6
The general solution of the second order DE
y'' − 3y' + 2y = 0
is
y = Ae2x + Bex
If we have the following boundary conditions:
y(0) = 4, y'(0) = 5
then the particular solution is given by:
y = e2x + 3ex
Now we do some examples using second order DEs. We are given a solution and we need to check if it is the correct solution.
Example 7 - Second Order DE
Show that
y = c1 sin 2x + 3 cos 2x
is a general solution for the differential equation
Example 8 - Second Order DE
Show that 
has a
solution of y = c1 + c2e2x
Didn't find what you are looking for on this page? Try search:
The IntMath Newsletter
Sign up for the free IntMath Newsletter. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!
Want Improved Memory?
Get a Memory... Be Phenomenal!
Book mark this page
Add this page to Del.icio.us, Furl, Digg, StumbleUpon, Google, whatever...
Need a break? Play a math game. Well, they all involve math... No, really!
Home |
Sitemap |
About & Contact |
Feedback & questions |
Privacy
Users online:
35. Highest ever: 170, on 16 November 2008
Page last modified: 05 June 2008
Valid HTML 4.01
| Valid CSS
SNB files