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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...
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Differential Equations - Laplace Transforms: Solving Initial Value Problems - i


It’s now time to get back to differential equations. We’ve spent the last three sections learning how to take Laplace transforms and how to take inverse Laplace transforms. These are going to be invaluable skills for the next couple of sections so don’t forget what we learned there.

Before proceeding into differential equations we will need one more formula. We will need to know how to take the Laplace transform of a derivative. First recall that f(n) denotes the nth derivative of the function f. We now have the following fact.


Fact


Suppose that f, f′, f″,… f(n−1) are all continuous functions and f(n) is a piecewise continuous function. Then,
L{f(n)}=snF(s)−sn−1f(0)−sn−2f′(0)−⋯−sf(n−2)(0)−f(n−1)(0)

Since we are going to be dealing with second order differential equations it will be convenient to have the Laplace transform of the first two derivatives.
L{y′}=sY(s)−y(0)L{y″}=s2Y(s)−sy(0)−y′(0)
Notice that the two function evaluations that appear in these formulas, y(0) and y′(0), are often what we’ve been using for initial condition in our IVP’s. So, this means that if we are to use these formulas to solve an IVP we will need initial conditions at t=0.
While Laplace transforms are particularly useful for nonhomogeneous differential equations which have Heaviside functions in the forcing function we’ll start off with a couple of fairly simple problems to illustrate how the process works.


Example 1 Solve the following IVP.y″−10y′+9y=5t,y(0)=−1y′(0)=2
The first step in using Laplace transforms to solve an IVP is to take the transform of every term in the differential equation.
L{y″}−10L{y′}+9L{y}=L{5t}Using the appropriate formulas from our table of Laplace transforms gives us the following.
s2Y(s)−sy(0)−y′(0)−10(sY(s)−y(0))+9Y(s)=5s2Plug in the initial conditions and collect all the terms that have a Y(s) in them.
(s2−10s+9)Y(s)+s−12=5s2Solve for Y(s).
Y(s)=5s2(s−9)(s−1)+12−s(s−9)(s−1)At this point it’s convenient to recall just what we’re trying to do. We are trying to find the solution, y(t), to an IVP. What we’ve managed to find at this point is not the solution, but its Laplace transform. So, in order to find the solution all that we need to do is to take the inverse transform.
Before doing that let’s notice that in its present form we will have to do partial fractions twice. However, if we combine the two terms up we will only be doing partial fractions once. Not only that, but the denominator for the combined term will be identical to the denominator of the first term. This means that we are going to partial fraction up a term with that denominator no matter what so we might as well make the numerator slightly messier and then just partial fraction once.
This is one of those things where we are apparently making the problem messier, but in the process we are going to save ourselves a fair amount of work!
Combining the two terms gives,
Y(s)=5+12s2−s3s2(s−9)(s−1)The partial fraction decomposition for this transform is,
Y(s)=As+Bs2+Cs−9+Ds−1Setting numerators equal gives,
5+12s2−s3=As(s−9)(s−1)+B(s−9)(s−1)+Cs2(s−1)+Ds2(s−9)Picking appropriate values of s and solving for the constants gives,
s=05=9B⇒B=59s=116=−8D⇒D=−2s=9248=648C⇒C=3181s=245=−14A+434581⇒A=5081Plugging in the constants gives,
Y(s)=5081s+59s2+3181s−9−2s−1Finally taking the inverse transform gives us the solution to the IVP.
y(t)=5081+59t+3181e9t−2et

That was a fair amount of work for a problem that probably could have been solved much quicker using the techniques from the previous chapter. The point of this problem however, was to show how we would use Laplace transforms to solve an IVP.

There are a couple of things to note here about using Laplace transforms to solve an IVP. First, using Laplace transforms reduces a differential equation down to an algebra problem. In the case of the last example the algebra was probably more complicated than the straight forward approach from the last chapter. However, in later problems this will be reversed. The algebra, while still very messy, will often be easier than a straight forward approach.
Second, unlike the approach in the last chapter, we did not need to first find a general solution, differentiate this, plug in the initial conditions and then solve for the constants to get the solution. With Laplace transforms, the initial conditions are applied during the first step and at the end we get the actual solution instead of a general solution.


In many of the later problems Laplace transforms will make the problems significantly easier to work than if we had done the straight forward approach of the last chapter. Also, as we will see, there are some differential equations that simply can’t be done using the techniques from the last chapter and so, in those cases, Laplace transforms will be our only solution.
Let’s take a look at another fairly simple problem.


Example 2 Solve the following IVP.2y″+3y′−2y=te−2t,y(0)=0y′(0)=−2
As with the first example, let’s first take the Laplace transform of all the terms in the differential equation. We’ll the plug in the initial conditions to get,
2(s2Y(s)−sy(0)−y′(0))+3(sY(s)−y(0))−2Y(s)=1(s+2)2(2s2+3s−2)Y(s)+4=1(s+2)2Now solve for Y(s).
Y(s)=1(2s−1)(s+2)3−4(2s−1)(s+2)Now, as we did in the last example we’ll go ahead and combine the two terms together as we will have to partial fraction up the first denominator anyway, so we may as well make the numerator a little more complex and just do a single partial fraction. This will give,
Y(s)=1−4(s+2)2(2s−1)(s+2)3=−4s2−16s−15(2s−1)(s+2)3The partial fraction decomposition is then,
Y(s)=A2s−1+Bs+2+C(s+2)2+D(s+2)3Setting numerator equal gives,
−4s2−16s−15=A(s+2)3+B(2s−1)(s+2)2+C(2s−1)(s+2)+D(2s−1)=(A+2B)s3+(6A+7B+2C)s2+(12A+4B+3C+2D)s+8A−4B−2C−DIn this case it’s probably easier to just set coefficients equal and solve the resulting system of equation rather than pick values of s. So, here is the system and its solution.
s3:A+2B=0s2:6A+7B+2C=−4s1:12A+4B+3C+2D=−16s0:8A−4B−2C−D=−15}⇒A=−192125B=96125C=−225D=−15We will get a common denominator of 125 on all these coefficients and factor that out when we go to plug them back into the transform. Doing this gives,
Y(s)=1125(−1922(s−12)+96s+2−10(s+2)2−252!2!(s+2)3)Notice that we also had to factor a 2 out of the denominator of the first term and fix up the numerator of the last term in order to get them to match up to the correct entries in our table of transforms.
Taking the inverse transform then gives,
y(t)=1125(−96et2+96e−2t−10te−2t−252t2e−2t)

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