Homogeneous differential equation, Mathematics

Assignment Help:

Assume that Y1(t) and Y2(t) are two solutions to (1) and y1(t) and y2(t) are a fundamental set of solutions to the associated homogeneous differential equation (2) so,

Y1 (t) - Y2 (t) is a solution to (2) and it can be written as

Y1 (t ) - Y2 (t ) =  c1 y1 (t ) + c2 y2 (t)

Note the notation used now. Capital letters considered as solutions to (1) while lower case letters considered as to solutions to (2. It is a fairly common convention while dealing with non-homogeneous differential equations.

This theorem is simple enough to prove thus let's do that. To prove this Y1(t) - Y2(t) is a solution to (2) all we require to do is plug this in the differential equation and check this.

(Y1 (t ) - Y2 (t ))'' + p(t) (Y1 (t ) - Y2 (t ))' + q(t) (Y1 (t ) - Y2 (t )) = 0

Y''1 + p(t) Y'1 + q(t) Y1 - (Y''2 + p(t) Y'2 + q(t) Y2) = 0

g(t) - g(t) = 0

0 = 0

We utilized the fact that Y1(t) and Y2(t) are two solutions to (1) into the third step. Since they are solutions to (1) we know as

Y''1 + p(t) Y'1 + q(t) Y1 = g(t)

Y''2 + p(t) Y'2 + q(t) Y2 = g(t)

Therefore, we were capable to prove that the difference of the two solutions is a solution to (2).

Proving as,

Y1 (t) - Y2 (t) = c1 y1 (t ) + c2 y2 (t) is even easier.

As y1(t) and y2(t) are a fundamental set of solutions to (2) we identify that they form a general solution and thus any solution to (2) can be written as,

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

Well, Y1(t) - Y2(t) is a solution to (2), as we've illustrated above, thus it can be written as,

Y1 (t) - Y2 (t) = c1 y1 (t ) + c2 y2 (t)

Thus, what does this theorem do for us? We can utilize this theorem to write down the type of the general solution to (1). Let's assume that y(t) is the general solution to (1) and that YP(t) is any solution to (1) which we can get our hands on. After that using the second part of our theorem as,

y(t) - Yp(t) = c1 y1 (t) + c2 y2(t)

Here y1(t) and y2(t) are a fundamental set of solutions for (2). So solving for y(t) provides,

y(t) = c1 y1 (t) + c2 y2(t) + Yp(t)

We can here call,

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

The complementary solution and YP(t) a specific solution. The general solution to a differential equation can after that be written as,

y(t) = yc + Yp(t)

Here, to solve a nonhomogeneous differential equation, we will require solving the homogeneous differential equation, (2), that for constant coefficient differential equations is pretty simple to do, and we'll require a solution to (1).

It seems to be a circular argument. So as to write down a solution to (1) we require a solution. Though, this isn't the problem that this seems to be. There are ways to get a solution to (1).

They just won't, in common, be the general solution. Actually, the next two sections are devoted to accurately that, finding a particular solution to a non-homogeneous differential equation.

There are two general methods for determining particular solutions: Undetermined Coefficients and Variation of Parameters. Both have their disadvantages and advantages as you will see in the subsequent couple of sections.


Related Discussions:- Homogeneous differential equation

Bernoulli differential equations, In this case we are going to consider dif...

In this case we are going to consider differential equations in the form, y ′ +  p   ( x ) y =  q   ( x ) y n Here p(x) and q(x) are continuous functions in the

What is the purpose of the reparameterisation, We have independent observat...

We have independent observations Xi, for i = 1, . . . , n, from a mixture of m Poisson distributions with component probabilities d c and rates l c, for c = 1, . . . ,m. We decid

Indices, 4n to the power 3/2 = 8 to the power minus 1/3. find the value of ...

4n to the power 3/2 = 8 to the power minus 1/3. find the value of n.

Solid mensuration, given dimensions: 130cm, 180cm, and 190cm is to be divid...

given dimensions: 130cm, 180cm, and 190cm is to be divided by a line bisecting the longest side shown from its opposite vertex. what''s the area adjacent to 180cm? ;

Equivalence class and equivalence relation, 1. For a function f : Z → Z, le...

1. For a function f : Z → Z, let R be the relation on Z given by xRy iff f(x) = f(y). (a) Prove that R is an equivalence relation on Z. (b) If for every x ? Z, the equivalenc

Full asymptotic expansion , Consider the integral where the notatio...

Consider the integral where the notation means a contour that is parallel to the real z axis, but moved down by a distance d . Use the method of steepest descents to deri

How many feet is the width of the deck, A pool is surrounded through a deck...

A pool is surrounded through a deck that has the similar width all the way around. The total area of the deck only is 400 square feet. The dimensions of the pool are 18 feet throug

Even and odd functions, Even and Odd Functions : This is the final topic ...

Even and Odd Functions : This is the final topic that we have to discuss in this chapter.  Firstly, an even function is any function which satisfies,

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