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Chapter 9: Problem 1

Show that \(y=x-x^{-1}\) is a solution of the differential equation \(x y^{\prime}+y=2 x\)

Short Answer

Expert verified
Yes, \(y = x - x^{-1}\) is a solution of the differential equation.

Step by step solution

01

Differentiate the Given Function

First, let's differentiate the function with respect to \(x\). Our function is \(y = x - x^{-1}\). Differentiating term by term, we get:\[y' = \frac{d}{dx}(x) - \frac{d}{dx}(x^{-1})\]This results in:\[y' = 1 + x^{-2}\]
02

Substitute into the Differential Equation

Substitute \(y = x - x^{-1}\) and \(y' = 1 + x^{-2}\) into the differential equation \(x y' + y = 2x\).This gives:\[x(1 + x^{-2}) + (x - x^{-1}) = 2x\]Simplifying this expression, we have:\[(x + x^{-1} + x - x^{-1}) = 2x\]This further simplifies to:\[2x = 2x\]which is true, confirming that \(y = x - x^{-1}\) satisfies the differential equation.

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus necessary for finding the rate at which a function changes. When dealing with problems that involve differential equations, understanding how to differentiate functions is crucial.

In the given problem, we started with the function \(y = x - x^{-1}\). Our task was to find \(y'\), the derivative of \(y\) with respect to \(x\). We applied the differentiation rule term by term:
  • For the first term, \(x\), the derivative is \(1\) because the derivative of \(x^n\) is \(nx^{n-1}\) when \(n=1\).
  • For the second term, \(x^{-1}\), the derivative is \(-x^{-2}\) because the derivative of \(x^{-n}\) is \(-nx^{-n-1}\).
Combining these results, we get \(y' = 1 + x^{-2}\). This derivative tells us how the function \(y\) changes with a small change in \(x\) and is an integral part of solving the given differential equation.
Substitution Method
The substitution method involves replacing variables in an equation with known expressions to transform the equation into a simpler form for solving. This technique is extensively used in solving differential equations, including the one in our exercise.

Once we found our derivative, \(y' = 1 + x^{-2}\), the next step required substituting both \(y\) and \(y'\) into the differential equation \(x y' + y = 2x\). This is how substitution works:
  • Replace \(y'\) with \(1 + x^{-2}\) in the equation.
  • Replace \(y\) with \(x - x^{-1}\) in the same equation.
  • Combine: \(x (1 + x^{-2}) + (x - x^{-1}) = 2x\).
By making these substitutions, the equation simplifies significantly, transforming a potentially complex equation into an easily verifiable identity.
Algebraic Manipulation
Algebraic manipulation is the practice of rearranging equations and expressions to reach a desired form or solution. In the context of differential equations, it often follows differentiation and substitution.

Following the substitution of \(y\) and \(y'\) into the differential equation, we engaged in algebraic manipulation to simplify the expression:
  • First, expand the term \(x (1 + x^{-2})\) which results in \(x + x^{-1}\).
  • Add \(x - x^{-1}\) from the \(y\) substitution.
  • The expression \((x + x^{-1} + x - x^{-1})\) simplifies to \(2x\), proving the identity \(2x = 2x\).
This final algebraic manipulation shows that the original function \(y = x - x^{-1}\) is indeed a solution of the differential equation. Mastery of such manipulations is pivotal in solving and verifying solutions to differential equations efficiently.

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Most popular questions from this chapter

Find the orthogonal trajectories of the family of curves.Use a graphing device to draw several members of each family on a common screen. $$y^{2}=k x^{3}$$

A Bernoulli differential equation (named after James Bernoulli) is of the form $$\frac{d y}{d x}+P(x) y=Q(x) y^{n}$$ Observe that, if \(n=0\) or \(1,\) the Bernoulli equation is linear. For other values of \(n,\) show that the substitution \(u=y^{1-n}\) transforms the Bernoulli equation into the linear equation $$\frac{d u}{d x}+(1-n) P(x) u=(1-n) Q(x)$$

Suppose you have just poured a cup of freshly brewed coffee with temperature \(95^{\circ} \mathrm{C}\) in a room where the temperature is \(20^{\circ} \mathrm{C}\) . (a) When do you think the coffee cools most quickly? What happens to the rate of cooling as time goes by? Explain. (b) Newton's Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings, provided that this difference is not too large. Write a differential equation that expresses Newton's Law of Cooling for this particular situation. What is the initial condition? In view of your answer to part (a), do you think this differential equation is an appropriate model for cooling? (c) Make a rough sketch of the graph of the solution of the initial-value problem in part (b).

\(11-14\) Sketch the direction field of the differential equation. Then use it to sketch a solution curve that passes through the given point. $$y^{\prime}=y+x y, \quad(0,1)$$

\(21-22\) Solve the differential equation and use a graphing calculator or computer to graph several members of the family of solutions. How does the solution curve change as \(C\) varies? $$x y^{\prime}+2 y=e^{x}$$
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