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

Derivatives, What are the ingredients of a Mathematical Model? What is a mo...

What are the ingredients of a Mathematical Model? What is a model?

MATLAB, program of curve revolve and create a surface

program of curve revolve and create a surface

Math, weather prediction on monsoon past data on project

weather prediction on monsoon past data on project

Market orientation, what is market orientation? what is the importance of ...

what is market orientation? what is the importance of market orientation?what are its implementation?

Constant aceleration formulae, a car comes to a stop from a speed of 30m/s ...

a car comes to a stop from a speed of 30m/s in a distance of 804m. The driver brakes so as to produce a decelration of 1/2m per sec sqaured to begin withand then brakes harder to p

Sketch the graph of the derivative of this function f '( x), Below is the s...

Below is the sketch of a function f ( x ) . Sketch the graph of the derivative of this function f ′ ( x ) . Solution : At first glance it seems to an all however impossib

Constructions, Draw a line segment AB of length 4.4cm. Taking A as centre, ...

Draw a line segment AB of length 4.4cm. Taking A as centre, draw a circle of radius. 2cm and taking B as centre, draw another circle of radius 2.2cm. Construct tangents to each cir

HELP, HOW MANY TENS ONES AND HUNDRED ARE IN A GROUP OF 2

HOW MANY TENS ONES AND HUNDRED ARE IN A GROUP OF 2

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