Chapter 4: Q3E (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 $y (1) = Y_{o}, y' (1) = Y_{1}$ , where $Y_{o}$ and $Y_{1}$ are real constants.
$t^{2} y'' + y = cost$

Short Answer

Expert verified

The differential equation has a unique solution in $0 < t < \infty$ .

Step by step solution

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

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

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

So, $p (t) = 0, q (t) = \frac{1}{t^{2}}, g (t) = \frac{cost}{t^{2}}$

Check the result

Here p(t), q(t),g(t) is continuous functions in the interval $- \infty < t < 0 and 0 < t < \infty$ and the point $t_{0} = 1$ in the continuity interval $0 < t < \infty$ .

Therefore, the differential equation has a unique solution is $0 < t < \infty$ .

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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 $y (1) = Y_{o}, y' (1) = Y_{1}$ , where $Y_{o}$ and $Y_{1}$ are real constants.
$t^{2} y'' + y = cost$

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 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.t2y''+y=cost

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

Probability and Statistics

Statistics

Geometry

Applied Mathematics

Discrete Mathematics

Pure Maths

Study anywhere. Anytime. Across all devices.

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 $y (1) = Y_{o}, y' (1) = Y_{1}$ , where $Y_{o}$ and $Y_{1}$ are real constants.
$t^{2} y'' + y = cost$