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Chapter 8: Q42P (page 430)

Solve by use of Fourier series. Assume in each case that the right-hand side is a periodic function whose values are stated for one period.

y"+9y={x,0<x<10,-1<x<0

Short Answer

Expert verified

Answer

The solution of function y"+9y={x,0<x<10,-1<x<0is an=1π1-sinx-cosx.

Step by step solution

01

Given information from question

The function is given asy"+9y={x,0<x<10,-1<x<0.

02

The quadratic formula

The quadratic formula to find out the roots is,

-b±b2-4ac2a

03

Expand f(x) in a Fourier series

The given function is

y"+9y={x,0<x<10,-1<x<0

Hereis a function of period

fx={x,if0<x<10,if-1<x<0

The auxiliary equation is

D2+9=0D2=-9D=±3i

04

Calculate the complementary function

The complementary function is

yc=e3xAcos3x+Bsin3xfx=xif<x<10if-1<x<0an=1π∫11fxcosnxdan=1π∫100cosnxdx+∫1xcosnxdx

Solve equation further

an=1π∫01xcosnxdxan=1πxcosnx+nx01an=1π1-sinx-cosx

Thus, the solution of functionis.

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Most popular questions from this chapter

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

(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.

Use L32 and L11 to obtainL(t2sinat).

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.

14. Problem 8.

The momentum pof an electron at speednear the speedof light increases according to the formula p=mv1-v2c2, whereis a constant (mass of the electron). If an electron is subject to a constant force F, Newton’s second law describing its motion is localid="1659249453669" dpdx=ddxmv1-v2c2=F.

Find v(t)and show that v→cas t→∞. Find the distance travelled by the electron in timeif it starts from rest.

See all solutions

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