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Question: In Problems 1鈥10, determine all the singular points of the given differential equation.

9. (sin)y"-(in)y=0

Short Answer

Expert verified

The singular point exists in this differential equation for Q(x) is at =n for 0 where n=1,2,3.....

Step by step solution

01

Ordinary and Singular Points

A point isx0 called an ordinary point of equationy'+p(x) y'+q(x)y = 0 if both pand qare analytic at X 0 . Ifx0 is not an ordinary point, it is called a singular point of the equation

02

Find the singular points

The given differential equation is

(sin)y"-(in)y=0

Dividing the above equation by(sin) we get,

On comparing the above equation with y"+p(x)y'+q(x)y=0 , we find that,

Hence, p(x) and Q(x) are analytic except, perhaps, when their denominators are zero

.

For Q(X) this occurs at which implies

We see that Q(X) is actually analytic at which implies

Therefore, Q(x)is actually except at for

The singular point exists in this differential equation for Q(x) is at =n for 0 where n=1,2,3.....

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Most popular questions from this chapter

In Problems 1-10, use a substitution y=xr to find the general solution to the given equation for x>0.

d2y/dx2=1/x dy/dx-4/x2 y

Question 18: In Problems, find a power series expansion for f(X) , given the expansion for f(x).

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Aging spring. As a spring ages, its 鈥渟pring constant鈥 decreases on value. One such model for a mass-spring system with an aging spring is mx"(t)+bx'(t)+ke- 畏tx(t)=0 .

Where m is the mass, b the damping constant, k and 畏 positive constants and x(t) displacement of the spring from equilibrium position. Let m=1 kg, b=2 Nsec/m, k=1 N/m, 畏 =1 sec-1. The system is set in motion by displacing the mass 1m from it equilibrium position and releasing it (x(0)=1, x'(0)=0). Find at least the first four nonzero terms in a power series expansion of about t=0 of displacement.

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To derive the general solutions given by equations (17)- (20)for the non-homogeneous equation (16), complete the following steps.

(a) Substitutey(x)=n=0anxnand the Maclaurin series into equation (16)to obtain

(2a2-a0)+k=1[(k+2)(k+1)ak+2-(k+1)ak]xk=n=0(-1)n(2n+1)!x2n+1

(b) Equate the coefficients of like powers on both sides of the equation in part (a) and thereby deduce the equations

a2=a02,a3=16+a13,a4=a08,a5=140+a115,a6=a048,a7=195040+a1105

(c) Show that the relations in part (b) yield the general solution to (16)given in equations (17)-(20).

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