Problem 2 In Exercises $1-4,$ show that ... [FREE SOLUTION]

Chapter 7: Problem 2

In Exercises $1-4,$ show that each function $y=f(x)$ is a solution of the accompanying differential equation. $$ y^{\prime}=y^{2} $$ $$\text { a. }\quad y=-\frac{1}{x} \quad \text { b. } y=-\frac{1}{x+3} \quad \text { c. } y=-\frac{1}{x+C} $$

Short Answer

Expert verified

All functions satisfy the differential equation.

Step by step solution

Differentiate the function for part a

The given function is $ y = -\frac{1}{x} $. We need to find $ y' $, the derivative with respect to $ x $. Using the power rule and chain rule, we obtain:\[ y' = \frac{d}{dx}(-x^{-1}) = x^{-2} = \frac{1}{x^2} \]

Substitute and verify for part a

Now, we substitute $ y $ into $ y' = y^2 $ to check if the equation holds:\[ y^2 = \left(-\frac{1}{x}\right)^2 = \frac{1}{x^2} \]Since $ y' = \frac{1}{x^2} = y^2 $, $ y = -\frac{1}{x} $ satisfies $ y' = y^2 $.

Differentiate the function for part b

The function is $ y = -\frac{1}{x+3} $. Find $ y' $:\[ y' = \frac{d}{dx}\left(-\frac{1}{x+3}\right) = \frac{1}{(x+3)^2} \]

Substitute and verify for part b

Substitute $ y $ into $ y' = y^2 $:\[ y^2 = \left(-\frac{1}{x+3}\right)^2 = \frac{1}{(x+3)^2} \]Since $ y' = \frac{1}{(x+3)^2} = y^2 $, $ y = -\frac{1}{x+3} $ satisfies $ y' = y^2 $.

Differentiate the function for part c

The function is $ y = -\frac{1}{x+C} $. Calculate $ y' $:\[ y' = \frac{d}{dx}(-\frac{1}{x+C}) = \frac{1}{(x+C)^2} \]

Substitute and verify for part c

Substitute $ y $ into $ y' = y^2 $:\[ y^2 = \left(-\frac{1}{x+C}\right)^2 = \frac{1}{(x+C)^2} \]Since $ y' = \frac{1}{(x+C)^2} = y^2 $, $ y = -\frac{1}{x+C} $ satisfies $ y' = y^2 $.

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Key Concepts

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

Derivative

The concept of a derivative is central to calculus and describes how a function changes as its input changes. In simpler terms, the derivative tells us the rate at which a function is increasing or decreasing at any given point. Let's break this down further:

The derivative of a function, often denoted as $y'$ or $ \frac{dy}{dx} $, signifies the slope of the function at a point, resembling the steepness of a hill.
Computing the derivative involves applying rules like the power rule or the chain rule, which we'll explore next.
In the exercise, finding $y'$ is crucial to verify that it satisfies the differential equation $y' = y^2$.

So, when we derive $y = -\frac{1}{x}$, we apply these concepts to find $y' = \frac{1}{x^2}$, exactly as needed to match $y^2$. Understanding derivatives allows us to assess how well a function meets specific conditions, like in differential equations.

Power Rule

The power rule is one of the simplest and most popular rules for finding derivatives when dealing with functions of the form $x^n$. It states that if $y = x^n$, then the derivative $y'$ is $nx^{n-1}$.

The power rule makes it easy to handle functions where variables are raised to powers, simplifying the differentiation process significantly.
For example, in the exercise, to differentiate $y = -x^{-1}$, the power rule helps: the derivative $y'$ becomes $-1x^{-2} = \frac{1}{x^2}$.

Using the power rule efficiently forms the basis for tackling more complex problems and is instrumental in working through most calculus problems, including differential equations as seen here.

Chain Rule

The chain rule is crucial when differentiating composite functions, which involve a function within another function. The chain rule helps us differentiate such complicated functions with ease.

The chain rule states that if you have a function $g(x)$ inside another function $f(x) = h(g(x))$, then the derivative $f'(x)$ is given by $h'(g(x)) \cdot g'(x)$.
In our exercise scenarios, such as differentiating $y = -\frac{1}{x+3}$, the chain rule aids in handling the composite nature of the expression.
This rule combines with other rules like the power rule, as shown, to unpick complex differentiation tasks, resulting in an effective way to confirm solutions to equations like $y' = y^2$.

Mastery of the chain rule along with the power rule equips students with powerful tools for differential calculus, facilitating the investigation of intricate problems like those in differential equations focused on changes within changes.

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91影视

In Exercises \(1-4,\) show that each function \(y=f(x)\) is a solution of the accompanying differential equation. $$ y^{\prime}=y^{2} $$ $$\text { a. }\quad y=-\frac{1}{x} \quad \text { b. } y=-\frac{1}{x+3} \quad \text { c. } y=-\frac{1}{x+C} $$

Short Answer

Step by step solution

Differentiate the function for part a

Substitute and verify for part a

Differentiate the function for part b

Substitute and verify for part b

Differentiate the function for part c

Substitute and verify for part c

Key Concepts

Derivative

Power Rule

Chain Rule

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