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Consider the \(n\)th order differential equation $$ y^{(n)}-a y=0, $$ where \(a\) is a real number. In each exercise, some information is presented about the solutions of this equation. Use the given information to deduce both the order \(n(n \geq 1)\) of the differential equation and the value of the constant \(a\). (If more than one answer is $$ y(t)=t^{3} \text { is a solution of the differential equation. } $$

Short Answer

Expert verified
Question: Determine the order of the given differential equation and the value of the constant a, given that \(y(t) = t^3\) is a solution to the equation: \( y^{(n)} - ay = 0 \). Answer: The order of the given differential equation is \(n = 3\), and the value of the constant \(a\) is \(0\).

Step by step solution

01

Find the first few derivatives of the given solution

To begin, we will find the first few derivatives of the given solution, \(y(t) = t^3\): First derivative: $$ y'(t) = \frac{d}{dt} t^3 = 3t^2 $$ Second derivative: $$ y''(t) = \frac{d^2}{dt^2} t^3 = 6t $$ Third derivative: $$ y^{(3)}(t) = \frac{d^3}{dt^3} t^3 = 6 $$ Fourth derivative: $$ y^{(4)}(t) = \frac{d^4}{dt^4} t^3 = 0 $$
02

Check which derivative satisfies the given differential equation

We will plug the given solution and its derivatives into the equation until we find the derivative that satisfies the equation: For n = 1: $$ 3t^2 - at^3 \neq 0 $$ For n = 2: $$ 6t - a(3t^2) \neq 0 $$ For n = 3: $$ 6 - a(t^3) = 0 $$ We can see that for n=3, the equation can be satisfied.
03

Solve for the constant \(a\)

Now that we found the necessary derivative that satisfies the equation, we can solve for the constant \(a\): $$ 6 - a(t^3) = 0 $$ For the equation to hold true for every value of \(t\), \(a\) must be 0: $$ a = 0 $$ In conclusion, the given differential equation has an order of \(n = 3\), and the value of the constant \(a\) is \(0\). The differential equation can be written as: $$ y^{(3)}=0 $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equation Solutions
Understanding the solutions to differential equations is a fundamental aspect of calculus and mathematical modeling. A differential equation is an equation that involves derivatives of a function. The solution to a differential equation is a function or a set of functions that satisfies the equation.

In the context of the given exercise, the solution provided is a specific function, namely, the cubic function y(t) = t^3. Finding solutions involves taking derivatives of this function and comparing them with the original differential equation. In real-world scenarios, solutions to differential equations describe various phenomena, such as the motion of fluids, population growth, or the behavior of electrical circuits.

It is important for students to understand that not all differential equations have simple closed-form solutions, and sometimes numerical methods or approximations are employed to understand the behavior of the solutions.
Higher Order Derivatives
Higher order derivatives are essentially derivatives of derivatives. They provide deeper insights into the behavior of functions. For instance, if the first derivative of a function represents velocity, the second derivative represents acceleration. Similarly, higher order derivatives can be related to concepts like jerk and snap in physics.

In our exercise, the derivatives up to the third order were calculated for the function y(t) = t^3. With each step, we find the rate of change at an increasingly higher level of granularity. When we reach the fourth derivative and it equates to zero, this implies a certain predictable pattern in the function's behavior – it indicates that any further derivatives will also be zero. This characteristic leads us to identify the correct order of the differential equation when we match it to the given information.
Differential Equation Order
The order of a differential equation is determined by the highest derivative that appears in the equation. This is crucial because it reflects the level of complexity and the amount of information required to obtain a solution. An nth order differential equation requires n conditions to uniquely determine a solution.

As illustrated in the given problem, identifying the order of the equation was done by iterating through the derivatives of the given solution. Once we reached the third derivative, which was a constant, we could match it to the structure of the differential equation provided y^{(n)} - ay = 0, setting n equal to 3. This exercise not only shows how to find the order of a differential equation but also reinforces the concept that each order has unique implications in the physical interpretation of solutions.

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Most popular questions from this chapter

Concavity of the Solution Curve In the discussion of direction fields in Section 1.3, you saw how the differential equation defines the slope of the solution curve at a point in the ty-plane. In particular, given the initial value problem \(y^{\prime}=f(t, y), y\left(t_{0}\right)=\) \(y_{0}\), the slope of the solution curve at initial condition point \(\left(t_{0}, y_{0}\right)\) is \(y^{\prime}\left(t_{0}\right)=f\left(t_{0}, y_{0}\right)\). In like manner, a second order equation provides direct information about the concavity of the solution curve. Given the initial value problem \(y^{\prime \prime}=f\left(t, y, y^{\prime}\right), y\left(t_{0}\right)=\) \(y_{0}, y^{\prime}\left(t_{0}\right)=y_{0}^{\prime}\), it follows that the concavity of the solution curve at the initial condition point \(\left(t_{0}, y_{0}\right)\) is \(y^{\prime \prime}\left(t_{0}\right)=f\left(t_{0}, y_{0}, y_{0}^{\prime}\right)\). (What is the slope of the solution curve at that point?) Consider the four graphs shown. Each graph displays a portion of the solution of one of the four initial value problems given. Match each graph with the appropriate initial value problem. (a) \(y^{\prime \prime}+y=2-\sin t, \quad y(0)=1, \quad y^{\prime}(0)=-1\) (b) \(y^{\prime \prime}+y=-2 t, \quad y(0)=1, \quad y^{\prime}(0)=-1\) (c) \(y^{\prime \prime}-y=t^{2}, \quad y(0)=1, \quad y^{\prime}(0)=1\) (d) \(y^{\prime \prime}-y=-2 \cos t, \quad y(0)=1, \quad y^{\prime}(0)=1\)

For each differential equation, (a) Find the complementary solution. (b) Find a particular solution. (c) Formulate the general solution. $$ y^{(4)}-y=t+1 $$

(a) Obtain the general solution of the differential equation. (b) Impose the initial conditions to obtain the unique solution of the initial value problem. (c) Describe the behavior of the solution as \(t \rightarrow-\infty\) and \(t \rightarrow \infty\). In each case, does \(y(t)\) approach \(-\infty,+\infty\), or a finite limit? $$ 9 y^{\prime \prime}-6 y^{\prime}+y=0, \quad y(3)=-2, \quad y^{\prime}(3)=-\frac{5}{3} $$

Find the solution of the differential equation that satisfies the given conditions. $$ y^{(4)}-y=e^{-t}, \quad y(0)=0, \quad \lim _{t \rightarrow \infty} y(t)=0 $$

In each exercise, (a) Find the general solution of the differential equation. (b) If initial conditions are specified, solve the initial value problem. $$ y^{(4)}-y=0 $$

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