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Chapter 4: Q8E (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.

(1-t)y''+ty'-2y=sint;y(0)=1,y'(0)=1

Short Answer

Expert verified

The differential equation has a unique solution.

Step by step solution

01

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

The given differential equation is (1-t)y''+ty'-2y=sint.

It can be written asy''+t(1-t)y'-2(1-t)y=sint(1-t)

So,p(t)=t(1-t),q(t)=-2(1-t),g(t)=sint(1-t)

02

Check the result

From theorem (5) If p(t), q(t), and g(t) are continuous on an interval (a, b) that contains point t, then for any choice of the initial values Yo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 but shows discontinuity at t=1.

So, theorem (5) applies except for point t=1.

Since the initial conditions are not the point (1,0) but are at (0,1).

Therefore the differential equation has a unique solution.s

This is the required result.

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