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Find the general solution of the following differential equations (complementary function + particular solution). Find the particular solution by inspection or by (6.18),(6.23),or.(6.24), Alsofind a computer solution and reconcile differences if necessary, noticing especially whether the particular solution is in simplest form [see(6.26)and the discussionafter(6.15)].

(D2+1)y=8xsinx

Short Answer

Expert verified

The general solution given by differential equation is

y(x)=C1sinx+C2cosx+2x(sinxxcosx)

Step by step solution

01

Given data. 

Given equation is(D2+1)y=8xsinx

02

General solution of differential equation

A general solution to the nth order differential equation is one that incorporates a significant number of arbitrary constants. If one uses the variable approach to solve a first-order differential equation, one must insert an arbitrary constant as soon as integration is completed.

03

Find the general solution of given differential equation.(D2+1)y=8xsinx

The given differential equation is

(D2+1)y=8xsinx

The auxiliary equation can be written asm2+1=0

The roots of the equation are

data-custom-editor="chemistry" m=±i

The complementary function is

C.F=C1sinx+C2cosxP.Iy=xeix(Ax+B)y'=eix(B+iBx+A(2+ix)x)y''=eix(B(x2i)+A(x24ix2)

Hence,

eix(B(x2i)+A(x24ix2)=8xieix                                         A+iB+2iAx=8xieix

Solve the equation further,

A=2i,B=2yp=2xeix(1ix)yp=2x(sinxxcosx)

Hence the equation is

P.I=2x(sinxxcosx)C.S=C1sinx+C2cosx+2x(sinxxcosx)

The solution of the differential equation is

y(x)=C1sinx+C2cosx+2x(sinxxcosx)

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Most popular questions from this chapter

(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.

Using Problems 29 and 31b, show that equation (6.24) is correct.

Several Terms on the Right-Hand Side: Principle of Superposition So far we have brushed over a question which may have occurred to you: What do we do if there are several terms on the right-hand side of the equation involving different exponentials?

In Problem 33 to 38 , solve the given differential equations by using the principle of superposition [see the solution of equation (6.29) . For example, in Problem 33 , solve three differential equations with right-hand sides equal to the three different brackets. Note that terms with the same exponential factor are kept together; thus a polynomial of any degree is kept together in one bracket.

y"-5y'+6y=2ex+6x-5

Using (3.9), find the general solution of each of the following differential equations. Compare a computer solution and, if necessary, reconcile it with yours. Hint: See comments just after (3.9), and Example .

y+y/x2+1=1/(x+x2+1)

Consider an equation for damped forced vibrations (mechanical or electrical) in which the right-hand side is a sum of several forces or emfs of different frequencies. For example, in (6.32) let the right-hand side be F1eiω1t+F2eiω2t+F3eiω3t,

Write the solution by the principle of superposition. Suppose, for giventhat we adjust the system so that ω=ω1'; show that the principal term in the solution is then the first one. Thus, the system acts as a "filter" to select vibrations of one frequency from a given set (for example, a radio tuned to one station selects principally the vibrations of the frequency of that station).

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