Chapter 4: Q1E (page 210)

Show that if y(t) satisfies $y " - ty = 0$ , then y(-t) satisfies $y " + ty = 0$ .

Short Answer

Expert verified

Thus, it is proved that if $y (t)$ satisfies $y " - ty = 0$ , then $y (- t)$ satisfies $y " + ty = 0 .$

Step by step solution

General form

Chain rule of the derivative states that:

$\frac{d}{dx} [f (g (x))] = f' (g (x)) \times g' (x)$

Evaluate the given equation.

Given that $y (t)$ satisfies $y " - ty = 0$ .

To prove that $y (- t)$ satisfies $y " + ty = 0$ .

Let us assume that, $u (t) = y (- t)$ .

Using the chain rulefind the first and second derivatives of $u (t)$ .

Case (1):

$\begin{matrix} u' (t) = \frac{d}{dt} [y (- t)] \\ = y' (- t) \times (- t)' \\ = - y' (- t) \end{matrix}$

Case (2):

$\begin{matrix} u " (t) = [- y' (- t)]' \\ = - y " (- t) \times (- t)' \\ = y " (- t) \end{matrix}$

Substitute the values.

Substitute $t$ within $y " + ty = 0.$

$\begin{matrix} y " (- t) - (- t) y (- t) = 0 \\ y " (- t) + ty (- t) = 0 \end{matrix}$

Substitute the result of cases (1) and (2);

$u " (t) + tu (t) = 0$

Therefore, $u (t) = y (- t)$ satisfies the equation $y " + ty = 0$ .

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Show that if y(t) satisfies $y " - ty = 0$ , then y(-t) satisfies $y " + ty = 0$ .

Short Answer

Step by step solution

General form

Evaluate the given equation.

Substitute the values.

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Show that if y(t) satisfies y"-ty=0, then y(-t) satisfies y"+ty=0.

Short Answer

Step by step solution

General form

Evaluate the given equation.

Substitute the values.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

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Study anywhere. Anytime. Across all devices.

Show that if y(t) satisfies $y " - ty = 0$ , then y(-t) satisfies $y " + ty = 0$ .