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In the previous section we
looked at Bernoulli Equations and saw that in order to solve them we needed to
use the substitution 
. Upon using this substitution we were able to
convert the differential equation into a form that we could deal with (linear
in this case). In this section we want
to take a look at a couple of other substitutions that can be used to reduce
some differential equations down to a solvable form.
The first substitution we’ll take a look at will require the
differential equation to be in the form,
First order differential equations that can be written in
this form are called homogeneous
differential equations. Note that we
will usually have to do some rewriting in order to put the differential
equation into the proper form.
Once we have verified that the differential equation is a
homogeneous differential equation and we’ve gotten it written in the proper
form we will use the following substitution.
We can then rewrite this as,
and then remembering that both y and v are functions of x we can use the chain rule (recall that
is implicit differentiation from
Calculus I) to compute,
Under this substitution the differential equation is then,
As we can see with a small rewrite of the new differential
equation we will have a separable differential equation
after the substitution.
Let’s take a quick look at a couple of examples of this kind
of substitution.
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Example 1 Solve
the following IVP and find the interval of validity for the solution.
Solution
Let’s first divide both sides by to rewrite the differential equation as
follows,
Now, this is
not in the officially proper form as we have listed above, but we can see
that everywhere the variables are listed they show up as the ratio, y/x
and so this is really all the farther that we need to go. So, let’s plug the substitution into this
form of the differential equation to get,
Next, rewrite
the differential equation to get everything separated out.
Integrating
both sides gives,
We need to do a
little rewriting using basic logarithm properties in order to be able to
easily solve this for v.
Now
exponentiate both sides and do a little rewriting
Note that because c
is an unknown constant then so is and so we may as well just call this c as we did above.
Finally, let’s solve for v and then plug the substitution back in and we’ll play a little
fast and loose with constants again.
At this point
it would probably be best to go ahead and apply the initial condition. Doing that gives,
Note that we
could have also converted the original initial condition into one for in
terms of v and then applied it upon
solving the separable differential equation.
In this case however, it was probably a little easier to do it in
terms of y given all the logarithms
in the solution to the separable differential equation.
Finally, plug
in c and solve for y to get,
The initial
condition tells us that the “”
must be the correct sign and so the actual solution is,
For the
interval of validity we can see that we need to avoid and because we can’t allow negative numbers
under the square root we also need to require that,
So, we have two
possible intervals of validity,
and the initial
condition tells us that it must be .
The graph of
the solution is,
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Example 2 Solve
the following IVP and find the interval of validity for the solution.
Solution
On the surface this differential equation looks like it
won’t be homogeneous. However, with a
quick logarithm property we can rewrite this as,
In this form
the differential equation is clearly homogeneous. Applying the substitution and separating
gives,
Integrate both
sides and do a little rewrite to get,
You were able
to do the integral on the left right?
It used the substitution .
Now, solve for v and note that we’ll need to
exponentiate both sides a couple of times and play fast and loose with
constants again.
Plugging the
substitution back in and solving for y
gives,
Applying the
initial condition and solving for c
gives,
The solution is
then,
We clearly need
to avoid to avoid division by zero and so with the
initial condition we can see that the interval of validity is .
The graph of
the solution is,
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For the next substitution we’ll take a look at we’ll need
the differential equation in the form,
In these cases we’ll use the substitution,
Plugging this into the differential equation gives,
So, with this substitution we’ll be able to rewrite the
original differential equation as a new separable differential equation that we
can solve.
Let’s take a look at a couple of examples.
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Example 3 Solve
the following IVP and find the interval of validity for the solution.
Solution
In this case we’ll use the substitution.
Note that we
didn’t include the “+1” in our substitution.
Usually only the ax + by part gets included in the
substitution. There are times where
including the extra constant may
change the difficulty of the solution process, either easier or
harder, however in this case it doesn’t really make much difference so we
won’t include it in our substitution.
So, plugging
this into the differential equation gives,
As we’ve shown
above we definitely have a separable differential equation. Also note that to help with the solution
process we left a minus sign on the right side. We’ll need to integrate both sides and in
order to do the integral on the left we’ll need to use partial fractions. We’ll leave it to you to fill in the
missing details and given that we’ll be doing quite a bit of partial fraction
work in a few chapters you should really make sure that you can do the
missing details.
Note that we
played a little fast and loose with constants above. The next step is fairly messy but needs to
be done and that is to solve for v
and note that we’ll be playing fast and loose with constants again where we
can get away with it and we’ll be skipping a few steps that you shouldn’t
have any problem verifying.
At this stage
we should back away a bit and note that we can’t play fast and loose with
constants anymore. We were able to do
that in first step because the c
appeared only once in the equation. At
this point however the c appears
twice and so we’ve got to keep them around.
If we “absorbed” the 3 into the c
on the right the “new” c would be
different from the c on the left
because the c on the left didn’t
have the 3 as well.
So, let’s solve
for v and then go ahead and go back
into terms of y.
The last step
is to then apply the initial condition and solve for c.
The solution is
then,
Note that
because exponentials exist everywhere and the denominator of the second term
is always positive (because exponentials are always positive and adding a
positive one onto that won’t change the fact that it’s positive) the interval
of validity for this solution will be all real numbers.
Here is a graph
of the solution.
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Example 4 Solve
the following IVP and find the interval of validity for the solution.
Solution
Here is the substitution that we’ll need for this example.
&nbs |