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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 7, Problem 7.4.2b

In Exercises 1–4, show that each function y=f(x) is a solution of the accompanying differential equation.
2. y' = y²
b. y = -1/(x+3)

Verified step by step guidance
1
Start by identifying the given function: \(y = -\frac{1}{x+3}\).
Find the derivative of \(y\) with respect to \(x\). Use the chain rule to differentiate \(y = - (x+3)^{-1}\), which gives \(y' = -(-1)(x+3)^{-2} \cdot 1 = \frac{1}{(x+3)^2}\).
Express \(y^2\) by squaring the original function: \(y^2 = \left(-\frac{1}{x+3}\right)^2 = \frac{1}{(x+3)^2}\).
Compare the derivative \(y'\) and \(y^2\) to check if they are equal: both are \(\frac{1}{(x+3)^2}\).
Since \(y' = y^2\), conclude that the function \(y = -\frac{1}{x+3}\) satisfies the differential equation \(y' = y^2\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Differential Equations

A differential equation relates a function to its derivatives. Solving it means finding a function that satisfies this relationship. In this problem, verifying a solution involves checking if the given function and its derivative satisfy the equation y' = y².
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Derivative of a Function

The derivative represents the rate of change of a function with respect to its variable. To verify the solution, you must compute the derivative y' of the given function y = -1/(x+3) using differentiation rules, such as the chain or quotient rule.
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Substitution and Verification

After finding the derivative y', substitute both y and y' into the differential equation to check if the equality holds. This process confirms whether the given function is indeed a solution to the differential equation.
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