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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’t 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 company wishes to set aside funds for future expansion and so arranges to make continuous deposits into a savings account at the rate of \(\$ 10,000\) per year. The savings account earns \(5 \%\) interest compounded continuously. (a) Set up the differential equation that is satisfied by the amount \(f(t)\) of money in the account at time \(t.\) (b) Solve the differential equation in part (a), assuming that \(f(0)=0,\) and determine how much money will be in the account at the end of 5 years.

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