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*solving second order homogeneous differential equations*, here's the result:

A differential equation that can be written in the form
is called a first-order homogeneous linear differential equation. It's homogeneous because after placing all terms that include the unknown equation and its derivative on the left-hand side, the right-hand side is identically zero for all The expression What follows is the general solution of a first-order homogeneous linear differential equation. It might seem quite involved, but it is nothing more than one simple step after another. If you don't move on to a next step until you understand the current one, you will have no problem understanding the solution process. The fact that the process is somewhat involved is actually a good thing in our context; once you know how its done, and then see how long Maple takes to do it, you'll be convinced that Maple provides you with tools that really save a lot of time and brain power. Start with Then
and Remembering the derivative of the ln function, and using the chain rule, Thus, by substitution, Note that we have transformed the original equation into the form discussed at the end of the introduction to differential equations. So at this point we simply need to integrate both sides of the equation.
Integration and differentiation being inverses of each other, this equation is the same as The constant comes from the constant produced by the integration of the left side of the equation, and then moving it to the right. Now we need to solve a logarithmic equation, and that involves taking exponents of both sides of the equation.
This equation is the same as The factor is a constant, so substitute into the equation.
We have now found an equation that yields The bigger problem is that rather than knowing something about the absolute value of the unknown function, we want to say that
(Note: The section designated by the above horizontal line and the next one is provided for completeness. The above equation, with , is the general solution of a first-order homogeneous linear differential equation. While the following material is essential to justify the removal of the absolute value function, it is not essential to understanding how to execute the solution steps. However, there is one key point that you should know if you skip this section, and that is that the assumption that y(t) is a continuous function is essential to the justification for removing the absolute value function. Also, know that if you skip this section you are missing out on a good example of the use of some basic theorems of calculus to make an interesting argument. to skip the justification and continue with the rest of this introduction to first-order homogeneous linear differential equations.) To justify this step, first make the following change (for reasons you'll see in a second, let's revert back to ).
Now . Thus, by definition of the absolute value function, A property of the absolute value function says that . Hence, By assumption, Now the equation
says that the absolute value of the function is a constant function of
or that (Note that this conclusion is not necessarily true if the function inside the absolute value function is not continuous. Why? Hint: Define the function In either case, it is certainly true that we can write
where or , depending on which of the above is the case. Moving back to the other side of the equation yields the desired result.
Finally, this is the general solution of the homogeneous linear differential equation. Note that
An example: Solve for Solution:
becomes
with Thus, by the formula we just developed, Which evaluates to
or As a check, substitute
Which evaluates to
Another example: Solve for Solution:
has Now And, Thus, This simplifies to Checking our work,
These two examples illustrate how to find the general solution of a homogeneous linear differential equation. However, in most real world situations you generally want one specific solution that depends on certain initial conditions. In these cases you know, or are given, a value at some initial time, . The initial-value, , enables you to narrow down the general solution to a single specific solution. Basically, it causes the replacement of the arbitrary constant, Its solution is as follows. After transforming it into a form that we can integrate, integrate both sides between and
becomes And this becomes Now take the exponential of both sides.
Following a processes similar to the earlier solution of the absolute value enigma, either
or To decide this one, note that we already know the value of Thus, the second option is not possible, and this implies that Isolate
A couple of examples: Solve Solution: According to the formula, This evaluates to As a check, Which of course simplifies to
Here's another way to solve this kind of initial-value problem. Solve Solution: First solve the equation without concern for the initial condition.
Now solve for
Thus, For this initial condition This simplifies to A check, becomes which simplifies to The last section that you'll want to read before returning to any polite comments, suggestions, or corrections. |