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

Short Answer

Expert verified
All three functions are solutions to the differential equation \( y' = y^2 \).

Step by step solution

01

Differentiate the given functions

Let's find the derivative of each proposed solution function **y** with respect to **x**. This will help us verify if each function satisfies the differential equation. **a.** For \( y = -\frac{1}{x} \), the derivative is \( y' = \frac{1}{x^2} \).**b.** For \( y = -\frac{1}{x+3} \), the derivative is \( y' = \frac{1}{(x+3)^2} \).**c.** For \( y = -\frac{1}{x+C} \), the derivative is \( y' = \frac{1}{(x+C)^2} \).
02

Substitute into the differential equation

Substitute each expression for \( y \) and \( y' \) into the differential equation \( y' = y^2 \) to verify their equality.**a.** Substitute into \( y' = y^2 \):\( \frac{1}{x^2} = \left(-\frac{1}{x}\right)^2 = \frac{1}{x^2} \). Equality holds.**b.** Substitute into \( y' = y^2 \):\( \frac{1}{(x+3)^2} = \left(-\frac{1}{x+3}\right)^2 = \frac{1}{(x+3)^2} \). Equality holds.**c.** Substitute into \( y' = y^2 \):\( \frac{1}{(x+C)^2} = \left(-\frac{1}{x+C}\right)^2 = \frac{1}{(x+C)^2} \). Equality holds.

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Key Concepts

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

Solution Verification
When solving differential equations, verifying a solution is crucial to ensure correctness. This is done by checking whether a proposed function indeed satisfies the original differential equation. The process typically involves two main steps:

  • Take the derivative of the proposed function, if necessary.
  • Substitute the function and its derivative back into the differential equation to see if both sides are equal.
In our exercise, we were given a differential equation \( y' = y^2 \) and asked to verify if certain functions are solutions. For verification, once we found the derivatives of the functions, we substituted them alongside the functions into the differential equation. In each case, equality holds, proving that these functions solve the given equation.

Solution verification is a foundational technique in differential equations, ensuring only correct functions are recognized as solutions.
Derivatives
Understanding derivatives is key in working with differential equations. A derivative is the rate at which a function changes at any given point, typically denoted as \( y' \). It provides the necessary link between a function and the solutions to its related differential equations.

For example, consider the function \( y = -\frac{1}{x} \). Its derivative is calculated as \( y' = \frac{1}{x^2} \). Similarly, for functions like \( y = -\frac{1}{x+3} \) and \( y = -\frac{1}{x+C} \), the derivatives are \( y' = \frac{1}{(x+3)^2} \) and \( y' = \frac{1}{(x+C)^2} \), respectively.

Computing derivatives helps simplify the task of checking which functions satisfy given differential equations. Understanding the manipulation of derivatives, particularly within fractions or complex functions, is crucial for this process.

Practicing derivation strengthens your ability to solve not just single problems, but broad applications in calculus and beyond.
Function Substitution
Function substitution is a technique used to verify solutions of differential equations. In function substitution, we plug the proposed function and its derivative back into the differential equation. This checks if the left-hand side equals the right-hand side of the equation.

Take, for example, our differential equation \( y' = y^2 \). For each function, such as \( y = -\frac{1}{x} \), after deriving the expression \( y' = \frac{1}{x^2} \), we substitute \( y \) and \( y' \) back into the equation:
\( \frac{1}{x^2} = \left(-\frac{1}{x}\right)^2 = \frac{1}{x^2} \).

Function substitution effectively bridges the calculated derivatives and the algebraic expressions of the functions. By ensuring this equality holds, we confirm the function as a valid solution.

Supporting your solutions with function substitution not only demonstrates accuracy but builds confidence in comprehending the intricate nature of differential equations.

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Most popular questions from this chapter

This last equation is linear in the (unknown) dependent variable u. Solve the differential equations. \(x^{2} y^{\prime}+2 x y=y^{3}\)

Use a CAS to explore graphically each of the differential equations in Exercises \(21-24 .\) Perform the following steps to help with your explorations. a. Plot a slope field for the differential equation in the given \(x y-\) window. b. Find the general solution of the differential equation using your CAS DE solver. c. Graph the solutions for the values of the arbitrary constant \(C=-2,-1,0,1,2\) superimposed on your slope field plot. d. Find and graph the solution that satisfies the specified initial condition over the interval \([0, b] .\) e. Find the Euler numerical approximation to the solution of the initial value problem with 4 subintervals of the \(x\) -interval and plot the Euler approximation superimposed on the graph produced in part (d). f. Repeat part (e) for \(8,16,\) and 32 subintervals. Plot these three Euler approximations superimposed on the graph from part (e). g. Find the error \((y \text { exact })-y(\text { Euler })\) at the specified point \(x=b\) for each of your four Euler approximations. Discuss the improvement in the percentage error. $$ \begin{array}{l}{y^{\prime}=x+y, \quad y(0)=-7 / 10 ; \quad-4 \leq x \leq 4, \quad-4 \leq y \leq 4} \\ {b=1}\end{array} $$

In Exercises 31 and \(32,\) obtain a slope field and graph the particular solution over the specified interval. Use your CAS DE solver to find the general solution of the differential equation. $$ y^{\prime}=(\sin x)(\sin y), \quad y(0)=2 ; \quad-6 \leq x \leq 6, \quad-6 \leq y \leq 6 $$

In Exercises \(13-18\) , find the orthogonal trajectories of the family of curves. Sketch several members of each family. $$ y=c e^{-x} $$

Solve the differential equation in Exercises \(9-18\). $$ \frac{d y}{d x}=x^{2} \sqrt{y}, \quad y>0 $$

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