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Chapter 4: Q7E (page 199)

In Problems 5 through 8, determine whether Theorem 5 applies. If it does, then discuss what conclusions can be drawn. If it does not, explain why.

y''+ty'-t2y=0;y(0)=1,y(1)=0

Short Answer

Expert verified

The differential equation has no unique solution.

Step by step solution

01

Find the value of p(t),q(t),g(t)

The given differential equation is y''+ty'-t2y=0.

So,p(t)=t,q(t)=-t2,g(t)=0

02

Check the result

From theorem (5) If p(t), q(t), and g(t) are continuous on an interval (a,b) that contains the point t, then for any choice of the initial values Yo鈥�鈥�and鈥�鈥�Y1, there exists a unique solution y(1) on the same interval (a, b) to the initial value problems.

Here p(t),q(t),g(t) is a continuous function in the interval -鈭�<t<鈭�,but initial conditions are y(0)=1,y(1)=0.So it is not an initial value problem.

Therefore, the differential equation has no unique solution.

This is the required result.

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Most popular questions from this chapter

Find the solution to the initial value problem.

y''(x)-y'(x)-2y(x)=cosx-sin2x;鈥夆赌夆赌夆赌夆赌�y(0)=-720,鈥夆赌夆赌夆赌�y'(0)=15

Using the mass-spring analogy, predict the behavior as t鈫�鈭�of the solution to the given initial value problem. Then confirm your prediction by actually solving the problem.

(a).鈥�y''+16y=0;y(0)=2,y'(0)=0(b).鈥�y''+100y'+y=0;y(0)=1,y'(0)=0(c).鈥�y''-6y'+8y=0;y(0)=1,y'(0)=0(d).鈥�y''+2y'-3y=0;y(0)=-2,y'(0)=0(e).鈥�y''-y'-6y=0;y(0)=1,y'(0)=1

Find a general solution y''-3y'-11y=0

Find a particular solution to the given higher-order equation. y'''+y''-2y=tet+1

Solve the given initial value problem. w''-4w'+2w=0;w(0)=0,w'(0)=1
See all solutions

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