Variation of parameters, Mathematics

Assignment Help:

In the previous section we looked at the method of undetermined coefficients for getting a particular solution to

p (t) y′′ + q (t) y′ + r (t) y = g (t)    .....................(1)

and we observed that while it decreased things down to just an algebra problem, the algebra could become fairly messy. On the top of such undetermined coefficients will merely work for a fairly tiny class of functions.

The method of Variation of Parameters is a very general method which can be used in many more cases. Though, there are two drawbacks to the method. Firstly, the complementary solution is absolutely needed to do the problem. It is in contrast to the method of undetermined coefficients where this was advisable to contain the complementary solution on hand, although was not needed. Second, as we will notice, so as to complete the method we will be doing a couple of integrals and there is no guarantee as we will be capable to do the integrals. Therefore, while this will always be possible to write down a formula to find the particular solution, we may not be capable to actually get it if the integrals are too difficult or if we are not capable to find the complementary solution.

We're intended for derive the formula for variation of parameters. We'll start off through acknowledging as the complementary solution to (1) is

yc (t ) = c1 y1 (t ) + c2 y2 (t)

Remember also that it is the general solution to the homogeneous differential equation.

p (t ) y′′ + q (t ) y′ + r (t ) y = 0    ............. (2)

 As well as recall that so as to write down the complementary solution we know as y1(t) and y2(t) are a fundamental set of solutions.

What we're going to do is notice if we can determine a pair of functions, u1(t) and u2(t) hence,

YP (t) = u1(t) y1(t ) + u2(t ) y2(t )

It will be a solution to (1). We have two unknowns now and so we'll require two equations eventually. One equation is simple. Our proposed solution should satisfy the differential equation, thus we'll get the first equation by plugging our proposed solution in (1). The second equation can arrive from a variety of places. We are going to determine our second equation simply through making an assumption as we will make our work easier. We'll say more regarding to this shortly.

Therefore, let's start. If we're going to plug our given solution in the differential equation we're going to require some derivatives hence let's get those. The first derivative is,

YP (t) = u'1y1 + u1 y'1 + u'2 y2 + u2 y'2

There is the assumption. Just to create the first derivative easier to deal along with we are going to suppose that whatever u1(t) and u2(t) are they will satisfy the subsequent.

u'1 y1 + u'2 y2 = 0   .........................(3)

Here's no motive ahead of time to believe that it can be done. Though, we will notice that it will work out. We simply create this assumption on the hope which it won't cause problems down the road and to create the first derivative simple so don't get excited regarding to it.

With this assumption the first derivative turns into,

Y'P(t) = u'1 y1 + u'2 y2

The second derivative is after that,

Y''P(t) = u'1 y'1 + u1y''1 +u'2 y'2 + u2 y''2

Plug the solution and its derivatives in (1).

p(t) (u'1 y'1 + u1 y''1+ u2 y''2) + q(t) (u1 y'1 + u2 y'2) + r(t) (u1 y1 + u2 y2) = g(t)

 Rearranging a little gives the subsequent,

p(t) (u'1 y'1 + u'2 y'2) + u1(t) (p(t) y''1 + q(t) y'1 + r(t)y1) + u2(t) (p(t) y''2 + q(t) y'2 + r(t)y2) = g(t)

Here, both y1(t) and y2(t) are solutions to (2) and therefore the second and third terms are zero. Acknowledging this and rearranging a little provides us,

p(t) (u'1 y'1 + u'2 y'2) + u1(t)(0) + u2(t)(0) = g(t)

 (u'1 y'1 + u'2 y'2) = (g(t))/(p(t))   ............................(4)

We've mostly got the two equations which we need. Before proceeding we're try to go back and create a further assumption. The previous equation, (4), is in fact the one that we want, however, so as to make things simpler for us we are going to suppose that the function p(t) = 1.

Conversely, we are going to go back and start working along with the differential equation,

y′′ + q (t) y′ + r (t) y = g (t)

If the coefficient of the second derivative isn't one divide this out hence it turns into a one. The formula which we're going to be getting will suppose this! Upon doing this the two equations which we need so solve for the unknown functions are

 u1′ y1 + u2′ y2 = 0   ....................(5)

u1′ y1′ + u2′ y2′ = g (t )    ...........(6)

Notice that in this system we know the two solutions and therefore the simply two unknowns there are u1′ and u2′.  Solving this system is in fact quite easy. Firstly, solve (5) for u1′ and plug this in (6) and do several simplification.

u1′ = - (u2′ y2/y1)  ...................(7)

- (u2′ y2/y1) y'1 + u'2 y'2 = g(t)

u'2(y'2 - (y2y'1)/y1) = g(t)

u'2 =( y1g(t))/( y1 y'2-y2y'1)   ..................(8)

Thus, we now have an expression for u2′.  Plugging it in (7) will provide us an expression for u1'.

u'1 =- (( y2g(t))/( y1 y'2-y2y'1)).......(9)

Then, let's notice that,

W(y1y2) = y1y'2 - y'1y2≠ 0

Recall that y1(t) and y2(t) are a fundamental set of solutions and thus we identify that the Wronskian won't be zero!

At last, all that we require to do is integrate (8) and (9) so as to determine what u1(t) and u2(t) are.  Doing this provides,

951_Variation of Parameters.png

1476_Variation of Parameters1.png

Thus, provided we can do these integrals, an exact solution to the differential equation is

YP(t) = y1u1 + y2u2

1409_Variation of Parameters2.png

Therefore, let's summarize up what we've found here.


Related Discussions:- Variation of parameters

Solid Mensuration, The two sides of a triangle are 17cm and 28cm long, and ...

The two sides of a triangle are 17cm and 28cm long, and the length of the median drawn to the third side is equal to 19.5 cm. What is the distance from an endpoint of the median to

Regression, regression line drawn as Y=C+1075x, when x was 2, and y was 239...

regression line drawn as Y=C+1075x, when x was 2, and y was 239, given that y intercept was 11. calculate the residual

Example on abels theorem, Without solving, find out the Wronskian of two so...

Without solving, find out the Wronskian of two solutions to the subsequent differential equation. t 4 y'' - 2t 3 y' - t 8 y = 0 Solution : First thing that we want to d

Math, the size of my sitting room is 7metres by 6metres . i bought a rug fo...

the size of my sitting room is 7metres by 6metres . i bought a rug for covering the centre of its floor. one metre of the floor around the edge of the room is not to be covered by

Formulas of surface area - applications of integrals, Formulas of Surface A...

Formulas of Surface Area - Applications of integrals S = ∫ 2Πyds          rotation about x-axis S = ∫ 2Πxds          rotation about y-axis Where, ds = √ 1 + (1+ (dy /

Math World Problem, The ratio of gasoline to oil needed to run a chain-saw ...

The ratio of gasoline to oil needed to run a chain-saw is 16:1. If you have 3.5 mL of oil, how many millilitres of gasoline must you add to get the proper mixture?

What is the length of the longer base, The longer base of a trapezoid is th...

The longer base of a trapezoid is three times the shorter base. The nonparallel sides are congruent. The nonparallel side is 5 cm more that the shorter base. The perimeter of the t

Integration, sketch the curve y=9-x2 stating the coordinates of the turning...

sketch the curve y=9-x2 stating the coordinates of the turning point and of the intersections with the axes.

Rational and irrational numbers, RATIONAL NUMBERS All numbers of the ty...

RATIONAL NUMBERS All numbers of the type p/q where p and q are integer and q ≠0, are known as rational. Thus  it can be noticed that every integer is a rational number

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd