Question: Solve (12.12)andto get (12.15). Hint: Use Cramer's rule (Chapter 3, Section 3); note that the denominator determinant is the Wronskian [Chapter 3 , equation (8.5)) of the functionssin xand cos x.

Short Answer

Expert verified

The value ofAx' andBx' from equationAx'sinx'Bx'cosx'=0 andAx'cosx'Bx'sinx'=1 isAx'=cosx' is Bx'=sinx'.

Step by step solution

01

Given information

The given expressions are Ax'sinx'Bx'cosx'=0Ax'cosx'Bx'sinx'=1.

02

Definition of Green Function

In mathematics, a Green's function is the impulse response of an inhomogeneous linear differential operator defined on a domain with specified initial conditions or boundary conditions.

03

Solve the given function

Let's first rewrite the equations in a way:

Ax'sinx'Bx'cosx'=0Ax'cosx'+Bx'sinx'=1

Now we can use the Cramer's rule:

Ax'=0cosx'1sinx'sinx'cosx'cosx'sinx'=cosx'sin2x'+cos2x'=cosx'

Bx'=sinx'0cosx'1sinx'cosx'cosx'sinx'=sinx'

Notice that the denominator is the Wronskian of sinx'and cosx'because dsinx'/dx'=cosx' and dcosx'/dx'=sinx'.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Continuing the method used in derivingand, verify the Laplace transforms of higher-order derivatives ofgiven in the table (L35).

(a) Show that D(eaxy)=eax(D+a)y,D2(eaxy)=eax(D+a)2y, and so on; that is, for any positive integral n, Dn(eaxy)=eax(D+a)ny.

Thus, show that ifis any polynomial in the operator D, then L(D)(eaxy)=eaxL(D+a)y.

This is called the exponential shift.

(b) Use to show that (D-1)3(exy)=exD3y,(D2+D-6)(e-3xy)=e-3x(D2-5D)y..

(c) Replace Dby D-a, to obtain eaxP(D)y=P(D-a)eaxy

This is called the inverse exponential shift.

(d) Using (c), we can change a differential equation whose right-hand side is an exponential times a

polynomial, to one whose right-hand side is just a polynomial. For example, consider

(D2-D-6)y=10×e2x; multiplying both sides by e-3xand using (c), we get

e-3x(D2-D-6)y=[D+32-D+3-6"]"ye-3x=(D2+5D)ye-3x=10x

Show that a solution of (D2+5D)u=10xis u=x2-25x; then or use this method to solve Problems 23 to 26.

The speed of a particle on the x axis, x0, is always numerically equal to the square root of its displacement x. If x=0when t=0, find x as a function of t. Show that the given conditions are satisfied if the particle remains at the origin for any arbitrary length of time t0and then moves away; find x for t>t0for this case.

In Problems 13 to 15, find a solution (or solutions) of the differential equation not obtainable by specializing the constant in your solution of the original problem. Hint: See Example 3.

13. Problem 2

Find the orthogonal trajectories of each of the following families of curves. In each case, sketch or computer plot several of the given curves and several of their orthogonal trajectories. Be careful to eliminate the constant from y'for the original curves; this constant takes different values for different curves of the original family, and you want an expression for y'which is valid for all curves of the family crossed by the orthogonal trajectory you are trying to find. See equations (2.10)to (2.12)

y=kx2

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free