Chapter 4: Q5E (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.
$t^{2} z'' + tz' + z = cost; z (0) = 1, z' (0) = 0$

Short Answer

Expert verified

The differential equation has no unique solution in $- \infty < t < \infty$ .

Step by step solution

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

The given differential equation is $t^{2} z'' + tz' + z = cost$ .

It can be written as $z'' + \frac{1}{t} z' + \frac{1}{t^{2}} z = \frac{cost}{t^{2}}$ .

So, $p (t) = \frac{1}{t}, q (t) = \frac{1}{t^{2}}, g (t) = \frac{cost}{t^{2}}$

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 $Y_{o} and Y_{1}$ , there exists a unique solution y(1) on the same interval (a, b) to the initial value problems.

Here p(t),q(t), and g(t) are continuous functions in the interval $0 < t < \infty$ , and the point $t_{0} = 0$ isn't in the interval $- \infty < t < \infty$

Therefore, the differential equation has no unique solution in $- \infty < t < \infty$ .

This is the required result.

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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.
$t^{2} z'' + tz' + z = cost; z (0) = 1, z' (0) = 0$

Short Answer

Step by step solution

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

Check the result

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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.t2z''+tz'+z=cost;z(0)=1,z'(0)=0

Short Answer

Step by step solution

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

Check the result

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Calculus

Decision Maths

Mechanics Maths

Geometry

Pure Maths

Study anywhere. Anytime. Across all devices.

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.
$t^{2} z'' + tz' + z = cost; z (0) = 1, z' (0) = 0$