Differential Equations Example 5


Uploaded by TheIntegralCALC on 17.07.2010

Transcript:
Hi everyone! Welcome back to integralcalc.com. We're going to be doing another differential
equations problem today. This one is actually asking us to find the solutions of a differential
equations problem. They ask us to find explicit solutions if possible. And if explicit solutions
aren't possible to find, implicit solutions. So let's go ahead and dissect this problem.
The differential equation that we're given is dy over dx equals negative three y. And
we're given an initial condition of y of five equal to ten. Whenever we're solving differential
equations, and we have an initial condition like this, the first thing that we're always
gonna want to do is separate the variables on the function. x's on the one side and y's
on the other. We usually have y’s on the left and x's on the right. So we start by
multiplying both sides by dx. Then to continue separating variables, we
divide both sides by y. In this case, I'm just going to go ahead and divide by negative
three y to separate the y's from the x's. The second step in a differential equation
problem is to integrate both sides once we've separated these variables.
So now that we have our variables separated, we have our integrals simplified, we go ahead
and integrate. And integrating the left side will give us this negative one-third times
the integral of one over a variable, is natural log or ln of the absolute value of that variable.
And then of course we have to add c on the right side. We only add c to the right side
and we'll talk a little bit about c but it’s, c is confusing because it's an aggregator
of all of the constants in the problem. So we'll see it's going to end up sucking some
things together as we keep going on with this problem.
After we've separated variables, integrated both sides, now what we need to do is go ahead
and solve for y so that we can simplify all of these and then plug in our initial condition.
So to solve for y, let's go ahead and multiply both sides by negative three. That will get
rid of this coefficient here. So we'll be left with natural log of the absolute value
of y equals negative three x minus three c. Just go ahead and distribute that.
Then to get rid of the natural log here, ‘cause remember we're trying to solve for y, we need
to raise both sides of this equation to base e, in other words, the ln of y and the negative
three x minus three c become the exponents here on these e's. So we raise both sides
to base e because this e and natural log will then cancel out. So we're left with absolute
value of y equals e to the negative three x minus three c.
Let's go ahead now and simplify this exponent over here. What we need to do is use the formula
that tells us that when we have a polynomial exponent on e, this e to the negative three
x minus three c is the same thing as e to the negative three x times e to the negative
three c. You can break apart the exponent like that as long as you remember to keep
the negative signs, these two e’s with their exponents multiplied together and sort of
everything in one exponent. So you break those apart and then here's where
we get to the part about c that's rather difficult. So we need to treat c as a constant. c is
like a place holder. It’s not truly, you know, mathematical. It basically groups everything,
all of the constants in the function together, so in other words, it's representing a constant.
If you plug in a constant like the number two for c, you'll get e to the negative six
and e to the negative six would just be some constant number. I could do the math here,
it's point zero zero two four seven eight zero. So this is a constant.
So c, and it's not involved with the variable here x. This whole e to the negative three
c can be replaced by just c itself. So if we replace this with c, then c becomes a coefficient,
we have absolute value of y equals c e to the negative three x.
We can take the absolute value brackets off of y, whenever, we do that by adding positive
or negative to the side. y equals positive or negative c e to the  negative three x
but c, remember is an aggregator of those constants  so it doesn't matter whether it's
positive or a negative because we're going to end up solving for c, so we can just drop
this positive and negative here. We've gone ahead and solved for y. This is
our function. Now that we've done this, we can go ahead and plug in our initial condition,
5 for x and 10 for y. The purpose of doing this is eventually to solve for c. So when we've done that, we get
ten equals c e to the negative fifteen and to solve for c, let's go ahead and divide
both sides by e to the negative fifteen equals c. And because we have a negative sign on
this exponent, we can make it positive by moving this whole e to the negative fifteen
to the numerator. That will cause the exponent to flip from a negative to a positive. So
c is going to equal ten e to the fifteen. And now that we've solved for c, our last
and final step is to go ahead and our answer for c back into this function here, this equation
that we found. So now that we've plugged c back in, all we
need to do is simplify and we're done. So remember that rule before, when we broke apart
the two different terms that were in the exponent on the e and multiplied them together? We
can just reverse that and we'll get y equals ten e, and then we can combine these exponents,
since these two e's are multiplied together. And we'll get ten e to the negative three
x plus fifteen. And that will actually be our final answer.
And because we were able to solve for y and only have this y on the left side, this is
an explicit solution, as opposed to implicit solution. So that's it. Hope that helped.
I'll see you guys later. Bye!