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Calculus III - 3-Dimensional Space: Equations of Lines

In this section we need to take a look at the equation of a line in  R 3 R 3 . As we saw in the previous section the equation  y = m x + b y = m x + b  does not describe a line in  R 3 R 3 , instead it describes a plane. This doesn’t mean however that we can’t write down an equation for a line in 3-D space. We’re just going to need a new way of writing down the equation of a curve. So, before we get into the equations of lines we first need to briefly look at vector functions. We’re going to take a more in depth look at vector functions later. At this point all that we need to worry about is notational issues and how they can be used to give the equation of a curve. The best way to get an idea of what a vector function is and what its graph looks like is to look at an example. So, consider the following vector function. → r ( t ) = ⟨ t , 1 ⟩ r → ( t ) = ⟨ t , 1 ⟩ A vector function is a function that takes one or more variables, one in this case, and returns a vector. Note as we

Differential Equations - Second Order: Variation of Parameters - i






In the last section we looked at the method of undetermined coefficients for finding a particular solution to
(1)p(t)y+q(t)y+r(t)y=g(t)
and we saw that while it reduced things down to just an algebra problem, the algebra could become quite messy. On top of that undetermined coefficients will only work for a fairly small class of functions.
The method of Variation of Parameters is a much more general method that can be used in many more cases. However, there are two disadvantages to the method. First, the complementary solution is absolutely required to do the problem. This is in contrast to the method of undetermined coefficients where it was advisable to have the complementary solution on hand but was not required. Second, as we will see, in order to complete the method we will be doing a couple of integrals and there is no guarantee that we will be able to do the integrals. So, while it will always be possible to write down a formula to get the particular solution, we may not be able to actually find it if the integrals are too difficult or if we are unable to find the complementary solution.
We’re going to derive the formula for variation of parameters. We’ll start off by acknowledging that the complementary solution to (1) is
yc(t)=c1y1(t)+c2y2(t)
Remember as well that this is the general solution to the homogeneous differential equation.
(2)p(t)y+q(t)y+r(t)y=0
Also recall that in order to write down the complementary solution we know that y1(t) and y2(t) are a fundamental set of solutions.
What we’re going to do is see if we can find a pair of functions, u1(t) and u2(t) so that
YP(t)=u1(t)y1(t)+u2(t)y2(t)
will be a solution to (1). We have two unknowns here and so we’ll need two equations eventually. One equation is easy. Our proposed solution must satisfy the differential equation, so we’ll get the first equation by plugging our proposed solution into (1). The second equation can come from a variety of places. We are going to get our second equation simply by making an assumption that will make our work easier. We’ll say more about this shortly.
So, let’s start. If we’re going to plug our proposed solution into the differential equation we’re going to need some derivatives so let’s get those. The first derivative is
YP(t)=u1y1+u1y1+u2y2+u2y2
Here’s the assumption. Simply to make the first derivative easier to deal with we are going to assume that whatever u1(t) and u2(t) are they will satisfy the following.
(3)u1y1+u2y2=0
Now, there is no reason ahead of time to believe that this can be done. However, we will see that this will work out. We simply make this assumption on the hope that it won’t cause problems down the road and to make the first derivative easier so don’t get excited about it.
With this assumption the first derivative becomes.
YP(t)=u1y1+u2y2
The second derivative is then,
YP(t)=u1y1+u1y1+u2y2+u2y2
Plug the solution and its derivatives into (1).
p(t)(u1y1+u1y1+u2y2+u2y2)+q(t)(u1y1+u2y2)+r(t)(u1y1+u2y2)=g(t)
Rearranging a little gives the following.
p(t)(u1y1+u2y2)+u1(t)(p(t)y1+q(t)y1+r(t)y1)+u2(t)(p(t)y2+q(t)y2+r(t)y2)=g(t)
Now, both y1(t) and y2(t) are solutions to (2) and so the second and third terms are zero. Acknowledging this and rearranging a little gives us,
p(t)(u1y1+u2y2)+u1(t)(0)+u2(t)(0)=g(t)(4)u1y1+u2y2=g(t)p(t)
We’ve almost got the two equations that we need. Before proceeding we’re going to go back and make a further assumption. The last equation, (4), is actually the one that we want, however, in order to make things simpler for us we are going to assume that the function p(t)=1.
In other words, we are going to go back and start working with the differential equation,
y+q(t)y+r(t)y=g(t)
If the coefficient of the second derivative isn’t one divide it out so that it becomes a one. The formula that we’re going to be getting will assume this! Upon doing this the two equations that we want to solve for the unknown functions are
(5)u1y1+u2y2=0(6)u1y1+u2y2=g(t)
Note that in this system we know the two solutions and so the only two unknowns here are u1 and u2. Solving this system is actually quite simple. First, solve (5) for u1 and plug this into (6) and do some simplification.
(7)u1=u2y2y1(u2y2y1)y1+u2y2=g(t)u2(y2y2y1y1)=g(t)u2(y1y2y2y1y1)=g(t)(8)u2=y1g(t)y1y2y2y1
So, we now have an expression for u2. Plugging this into (7) will give us an expression for u1.
(9)u1=y2g(t)y1y2y2y1
Next, let’s notice that
W(y1,y2)=y1y2y2y10
Recall that y1(t) and y2(t) are a fundamental set of solutions and so we know that the Wronskian won’t be zero!

Finally, all that we need to do is integrate (8) and (9) in order to determine what u1(t) and u2(t) are. Doing this gives,
u1(t)=y2g(t)W(y1,y2)dtu2(t)=y1g(t)W(y1,y2)dt
So, provided we can do these integrals, a particular solution to the differential equation is
YP(t)=y1u1+y2u2=y1y2g(t)W(y1,y2)dt+y2y1g(t)W(y1,y2)dt
So, let’s summarize up what we’ve determined here.


Variation of Parameters

Consider the differential equation,
y+q(t)y+r(t)y=g(t)Assume that y1(t) and y2(t) are a fundamental set of solutions for
y+q(t)y+r(t)y=0Then a particular solution to the nonhomogeneous differential equation is,
YP(t)=y1y2g(t)W(y1,y2)dt+y2y1g(t)W(y1,y2)dt
Depending on the person and the problem, some will find the formula easier to memorize and use, while others will find the process used to get the formula easier. The examples in this section will be done using the formula.
Before proceeding with a couple of examples let’s first address the issues involving the constants of integration that will arise out of the integrals. Putting in the constants of integration will give the following.
YP(t)=y1(y2g(t)W(y1,y2)dt+c)+y2(y1g(t)W(y1,y2)dt+k)=y1y2g(t)W(y1,y2)dt+y2y1g(t)W(y1,y2)dt+(cy1+ky2)
The final quantity in the parenthesis is nothing more than the complementary solution with c1 = -c and c2 = k and we know that if we plug this into the differential equation it will simplify out to zero since it is the solution to the homogeneous differential equation. In other words, these terms add nothing to the particular solution and so we will go ahead and assume that c = 0 and k = 0 in all the examples.
One final note before we proceed with examples. Do not worry about which of your two solutions in the complementary solution is y1(t) and which one is y2(t). It doesn’t matter. You will get the same answer no matter which one you choose to be y1(t) and which one you choose to be y2(t).
Let’s work a couple of examples now.

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