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Chapter 10: Problem 8

Is the constant function \(f(t)=-4\) a solution of the differential equation \(y^{\prime}=t^{2}(y+4) ?\)

Short Answer

Expert verified
Yes, f(t) = -4 is a solution.

Step by step solution

01

Identify the Given Function

The given function is f(t) = -4.
02

Calculate the Derivative

The derivative of a constant function is 0. Therefore, the derivative of f(t) = -4 is f'(t) = 0.
03

Substitute into the Differential Equation

Substitute f(t) = -4 and f'(t) = 0 into the differential equation y' = t^2 (y + 4). We simplify to get: 0 = t^2 (-4 + 4).
04

Simplify the Equation

Simplify the right side: t^2 (-4 + 4) = t^2 (0) = 0.
05

Verify the Solution

Since both sides of the equation equal 0, the given function f(t) = -4 satisfies the differential equation y' = t^2 (y + 4).

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Key Concepts

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

Constant Functions
Before diving into the solution, it's important to understand what a constant function is. A constant function is a function that always returns a fixed value, regardless of the input. In mathematical terms, if we have a function \(f(t)\), and it is constant, this means: \ \(f(t) = C\) for any \(t\), where \(C\) is a constant number.
In our exercise, the constant function provided is \(f(t) = -4\). It does not change no matter what value of \(t\) we input. This property makes them straightforward and predictable.
Derivatives
One of the key steps in verifying if \(f(t) = -4\) is a solution to the given differential equation involves calculating its derivative. The derivative of a function measures how it changes with respect to its input. For constant functions, this becomes very simple.
When you take the derivative of any constant function \(f(t) = C\) where \(C\) is a constant number, the result is always zero: \ \(f'(t) = \frac{d}{dt}C = 0\).
This is because constants don鈥檛 change, thus their rate of change is zero. In our case, the derivative of \(f(t) = -4\) is \(\frac{d}{dt}(-4) = 0\). This zero derivative becomes crucial when substituting into the differential equation later.
Verification of Solutions
After calculating the derivative of the function, the next step is verification. This ensures that the constant function \(f(t) = -4\) fulfills the requirements of the differential equation \(y' = t^2(y + 4)\).
Substituting \(f(t) = -4\) and \(f'(t) = 0\) into the equation, we get: \ \(0 = t^2(-4 + 4)\).
Simplifying within the parentheses first, we have: \ \(-4 + 4 = 0\).
So the equation becomes: \ \(0 = t^2(0) = 0\).
Since both sides of the equation equal zero, the function \(f(t) = -4\) is indeed a valid solution to the differential equation.
Verification is a crucial step in differential equations as it confirms the correctness of a proposed solution.

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Most popular questions from this chapter

A certain piece of news is being broadcast to a potential audience of 200,000 people. Let \(f(t)\) be the number of people who have heard the news after \(t\) hours. Suppose that \(y=f(t)\) satisfies $$y^{\prime}=.07(200,000-y), \quad y(0)=10$$ Describe this initial-value problem in words.

A cool object is placed in a room that is maintained at a constant temperature of \(20^{\circ} \mathrm{C}\). The rate at which the temperature of the object rises is proportional to the difference between the room temperature and the temperature of the object. Let \(y=f(t)\) be the temperature of the object at time \(t\); give a differential equation that describes the rate of change of \(f(t)\).

Solving the differential equations that arise from modeling may require using integration by parts. [See formula (1).] After depositing an initial amount of \(\$ 10,000\) in a savings account that earns \(4 \%\) interest compounded continuously, a person continued to make deposits for a certain period of time and then started to make withdrawals from the account. The annual rate of deposits was given by \(3000-500 t\) dollars per year, \(t\) years from the time the account was opened. (Here, negative rates of deposits correspond to withdrawals.) (a) How many years did the person contribute to the account before starting to withdraw money from it? (b) Let \(P(t)\) denote the amount of money in the account, \(t\) years after the initial deposit. Find an initial-value problem satisfied by \(P(t)\). (Assume that the deposits and withdrawals were made continuously.)

Suppose that, once a sunflower plant has started growing, the rate of growth at any time is proportional to the product of its height and the difference between its height at maturity and its current height. Give a differential equation that is satisfied by \(f(t),\) the height at time \(t,\) and sketch the solution.

A certain drug is administered intravenously to a patient at the continuous rate of \(r\) milligrams per hour. The paticnt's body removes the drug from the bloodstream at a rate proportional to the amount of the drug in the blood, with constant of proportionality \(k=.5\) (a) Write a differential equation that is satisfied by the amount \(f(t)\) of the drug in the blood at time \(t\) (in hours). (b) Find \(f(t)\) assuming that \(f(0)=0 .\) (Give your answer in terms of \(r .)\) (c) In a therapeutic 2 -hour infusion, the amount of drug in the body should reach 1 milligram within 1 hour of administration and stay above this level for another hour. However, to avoid toxicity, the amount of drug in the body should not exceed 2 milligrams at any time. Plot the graph of \(f(t)\) on the interval \(1 \leq t \leq 2,\) as \(r\) varies between 1 and 2 by increments of \(.1 .\) That is, plot \(f(t)\) for \(r=1,1.1,1.2,1.3, \ldots . .2 .\) By looking at the graphs, pick the values of \(r\) that yield a therapeutic and nontoxic 2-hour infusion.
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