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Chapter 4: Q4E (page 199)

In Problems 1 through 4, use Theorem 5 to discuss the existence and uniqueness of a solution to the differential equation that satisfies the initial conditions y(1)=Yo,y'(1)=Y1, where Yoand Y1are real constants.

ety''-1t-3y'+y=lnt

Short Answer

Expert verified

The differential equation has a unique solution in 0<t<3.

Step by step solution

01

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

The given differential equation is ety''-1t-3y'+y=lnt.

It can be written as y''-y'et(t-3)+yet(t-3)=lntet(t-3).

So,p(t)=-1et(t-3),q(t)=1et(t-3),g(t)=lntet(t-3)

02

Check the result

Here p(t),q(t),g(t) are continuous functions in the interval 0<t<3鈥�and鈥�3<t<鈭�and the point t0=1in the continuity interval0<t<3

Therefore, the differential equation has a unique solution in 0<t<3.

This is the required result.

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Most popular questions from this chapter

Find the solution to the initial value problem.

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