How to solve heat equation by method of variation of parameter?
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the method of undetermined coefficients for finding a particular solution to
p(t)y′′+q(t)y′+r(t)y=g(t)(1)(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)(1) is
yc(t)=c1y1(t)+c2y2(t)yc(t)=c1y1(t)+c2y2(t)
Remember as well that this is the general solution to the homogeneous differential equation.
p(t)y′′+q(t)y′+r(t)y=0(2)(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)y1(t) and y2(t)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)u1(t) and u2(t)u2(t) so that
YP(t)=u1(t)y1(t)+u2(t)y2(t)YP(t)=u1(t)y1(t)+u2(t)y2(t)
will be a solution to (1)(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)(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
Y′P(t)=u′1y1+u1y′1+u′2y2+u2y′2YP′(t)=u1′y1+u1y1′+u2′y2+u2y2′
Here’s the assumption. Simply to make the first derivative easier to deal with we are going to assume that whatever u1(t)u1(t) and u2(t)u2(t) are they will satisfy the following.
u′1y1+u′2
the method of undetermined coefficients for finding a particular solution to
p(t)y′′+q(t)y′+r(t)y=g(t)(1)(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)(1) is
yc(t)=c1y1(t)+c2y2(t)yc(t)=c1y1(t)+c2y2(t)
Remember as well that this is the general solution to the homogeneous differential equation.
p(t)y′′+q(t)y′+r(t)y=0(2)(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)y1(t) and y2(t)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)u1(t) and u2(t)u2(t) so that
YP(t)=u1(t)y1(t)+u2(t)y2(t)YP(t)=u1(t)y1(t)+u2(t)y2(t)
will be a solution to (1)(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)(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
Y′P(t)=u′1y1+u1y′1+u′2y2+u2y′2YP′(t)=u1′y1+u1y1′+u2′y2+u2y2′
Here’s the assumption. Simply to make the first derivative easier to deal with we are going to assume that whatever u1(t)u1(t) and u2(t)u2(t) are they will satisfy the following.
u′1y1+u′2
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