/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 In Problems 17 through 26, first... [FREE SOLUTION] | 91Ó°ÊÓ

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In Problems 17 through 26, first verify that \(y(x)\) satisfies the given differential equation. Then determine a value of the constant \(C\) so that \(y(x)\) satisfies the given initial condition. Use a computer or graphing calculator (if desired) to sketch several typical solutions of the given differential equation, and highlight the one that satisfies the given initial condition. \(y^{\prime}+y \tan x=\cos x ; y(x)=(x+C) \cos x, y(\pi)=0\)

Short Answer

Expert verified
The solution satisfying the initial condition is \( y(x) = (x - \pi) \cos x \).

Step by step solution

01

Verify that y(x) satisfies the differential equation

Given the differential equation is \( y' + y \tan x = \cos x \) and the function \( y(x) = (x + C) \cos x \). First, find the derivative \( y' \). Use the product rule for differentiation: \( \frac{d}{dx}[(x + C) \cos x] = (x+C)' \cos x + (x+C)(\cos x)' = \cos x - (x + C) \sin x\). Substitute \( y \) into the differential equation:\[ y = (x + C) \cos x \]\[ y' = \cos x - (x + C) \sin x \]Substitute into the differential equation:\[ \cos x - (x + C) \sin x + (x + C) \cos x \tan x = \cos x \]This simplifies to \[ \cos x - (x + C) \sin x + (x + C) \sin x = \cos x \], verifying that the equation is satisfied.
02

Determine the value of constant C

We are given the initial condition \( y(\pi) = 0 \). Substitute \( x = \pi \) into the equation:\[ 0 = (\pi + C) \cos \pi \]Since \( \cos \pi = -1 \), this gives us:\[ 0 = (\pi + C)(-1) \]Solving for \( C \), we get:\[ \pi + C = 0\]\[ C = -\pi\]
03

Interpret the result for graphing

Now that we have \( C = -\pi \), the particular solution satisfying the initial condition is \( y(x) = (x - \pi) \cos x \). When plotting, you would graph similar solutions with different \( C \) values, emphasizing the one where \( C = -\pi \). This graph should pass through the given point \( (\pi, 0) \). This verifies both the solution to the differential equation and the condition \( y(\pi) = 0 \).

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Key Concepts

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

Initial Condition
When working with differential equations, an initial condition helps to identify a unique solution from a family of solutions. The initial condition is simply a specific value that the function must pass through. In our original exercise, the initial condition given was \( y(\pi) = 0 \). This means the solution must pass through the point \((\pi, 0)\). This constraint allows us to determine the value of the constant \( C \) in the solution. By substituting \( x = \pi \) and \( y = 0 \) into the general solution \( y(x) = (x+C) \cos x \), we find that \( C \) must be \( -\pi \) to satisfy the initial condition.
Product Rule
The product rule is a fundamental tool in calculus used to find derivatives of functions that are products of two simpler functions. When you have a function that can be expressed as the product of two factors, like \((x + C) \cos x \) in our exercise, you will use the product rule to differentiate it.The product rule is stated as: If \( u(x) \) and \( v(x) \) are two differentiable functions, then the derivative \( (uv)' \) is given by:
  • \( (uv)' = u'v + uv' \)
In the step-by-step solution, we apply the product rule to differentiate \( (x + C) \cos x \). It involves differentiating both the \((x + C)\) component and the \( \cos x\) component separately, then summing the respective terms to find the derivative \( y' \). This careful application confirms that \( y(x) \) satisfies the differential equation \( y' + y \tan x = \cos x \).
Graphing Solutions
Graphing solutions of differential equations allows us to visualize how different functions behave under varying conditions. In this exercise, after finding that \( C = -\pi \), the task involves plotting solutions for varying values of \( C \), then highlighting the particular graph where \( C = -\pi \).Using tools like a computer or graphing calculator helps to:
  • Visualize the entire family of solutions represented by the differential equation.
  • Confirm that the solution with \( C = -\pi \) passes through the point \((\pi, 0)\), as required by the initial condition.
Graphing gives insights into the behavior of the solution over a range of \( x \)-values and helps understand the impact of different constants on the overall shape of the function.
Integration Constants
Integration constants play a crucial role in solving differential equations. When you integrate a function, a constant of integration \( C \) is introduced because the indefinite integral of a function is a family of functions. This arbitrary constant accounts for the various solutions of an antiderivative.In our problem, after verifying the differential equation, the constant \( C \) specified a particular solution given an initial condition. Without this condition, \( C \) would remain unknown, and the solution would represent a broad family of functions. The value of the integration constant \( C \) becomes crucial when we impose initial conditions like \( y(\pi) = 0 \). This operation allows us to narrow down to the unique solution that truly satisfies both the differential equation and initial condition constraints, leading to a precise understanding of the system described by the equation.

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Most popular questions from this chapter

Verify that if \(c>0\), then the function defined piecewise by $$ y(x)=\left\\{\begin{array}{ll} 0 & \text { if } x^{2} \leqq c \\ \left(x^{2}-c\right)^{2} & \text { if } x^{2}>c \end{array}\right. $$ satisfies the differential equation \(y^{\prime}=4 x \sqrt{y}\) for all \(x\). Sketch a variety of such solution curves for different values of \(c\). Then determine (in terms of \(a\) and \(b\) ) how many different solutions the initial value problem \(y^{\prime}=4 x \sqrt{y}\), \(y(a)=b\) has.

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Find general solutions of the differential equations. Primes denote derivatives with respect to \(x\) throughout. $$ x^{2} y^{\prime}+2 x y=5 y^{3} $$

Find a general solution and any singular solutions of the differential equation dymyslashd \(x=y \sqrt{y^{2}-1}\). Determine the points \((a, b)\) in the plane for which the initial value problem \(y^{\prime}=y \sqrt{y^{2}-1}, y(a)=b\) has (a) no solution, (b) a unique solution, (c) infinitely many solutions.

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