/*! 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 17 Find the solution of the differe... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the solution of the differential equation that satisfies the given initial condition. $$y ^ { \prime } \tan x = a + y , y ( \pi / 3 ) = a , \quad 0 < x < \pi / 2$$

Short Answer

Expert verified
The solution is \( y = \frac{2a}{\sqrt{3}} \sin x - a \).

Step by step solution

01

Separate Variables

First, we need to separate the variables in the differential equation. The given equation is \( y' \tan x = a + y \). We can rewrite this as \( \frac{dy}{dx} = \frac{a + y}{\tan x} \). Now separate the variables to get: \( \frac{dy}{a+y} = \frac{dx}{\tan x} \).
02

Integrate Both Sides

Integrate both sides to find the solution. The left side becomes \( \int \frac{1}{a+y} \, dy \) which integrates to \( \ln|a+y| + C_1 \). The right side integrates as follows: \( \int \frac{1}{\tan x} \, dx = \int \cot x \, dx = \ln|\sin x| + C_2 \).
03

Solve for y

Equating the integrals gives us \( \ln|a+y| = \ln|\sin x| + C \) where \( C = C_2 - C_1 \). Exponentiating both sides, we have \( |a+y| = K|\sin x| \) for some constant \( K = e^C \). Solve for \( y \) to get \( y = K\sin x - a \).
04

Apply the Initial Condition

Use the initial condition \( y(\pi/3) = a \). Substitute into the equation: \( a = K\sin(\pi/3) - a \). Given \( \sin(\pi/3) = \frac{\sqrt{3}}{2} \), we solve \( a = K \frac{\sqrt{3}}{2} - a \) leading to \( K = \frac{2a}{\sqrt{3}} \).
05

Write Final Solution

Substitute \( K = \frac{2a}{\sqrt{3}} \) back into the formula for \( y \): \( y = \frac{2a}{\sqrt{3}} \sin x - a \). Simplify as needed for clarity.

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Key Concepts

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

Initial Conditions
In differential equations, initial conditions are essential to find a unique solution. They are specific values given at a certain point, often used to determine constants in the solution of the equation.
You can think of initial conditions as a way to "anchor" the solution to a specific point on the curve. This not only determines the shape of the solution but also ensures it’s the correct one among many possible solutions.
For our problem, the initial condition is given as \( y(\pi/3) = a \). We use this to find the constant \( K \) in our solution after integrating both sides. By substituting these conditions into the equation, we essentially solve for any arbitrary constants and pin it down to match the real-world scenario or particular instance given in the problem.
Integration Techniques
Integration techniques are crucial tools in solving differential equations. They help in finding the antiderivative which is necessary to solve equations by integration.
In our example, we encountered integrals such as \( \int \frac{1}{a+y} \, dy \) and \( \int \cot x \, dx \). These integrations involve knowing how to handle logarithmic functions. The first integral results in \( \ln|a+y| \), and the second one evaluates to \( \ln|\sin x| \).
Various techniques exist, including substitution, integration by parts, partial fractions, and trigonometric integration. Understanding which to use relies heavily on the form of the differential equation. Here, recognizing \( \cot x \) as the integration form \( \frac{\cos x}{\sin x} \) is a key step.
Variable Separation
Separating variables is a powerful method to solve first-order differential equations. It involves rearranging the equation so that all terms involving one variable appear on one side and all terms involving the other variable appear on the other side.
This technique allows us to integrate both sides independently. In the given example, we start with the equation \( y' \tan x = a + y \).
We rearrange to get \( \frac{dy}{a+y} = \frac{dx}{\tan x} \), isolating terms with \( y \) on one side and terms with \( x \) on the other. This logical separation facilitates straightforward integration and leads us to form integrable expressions.
Trigonometric Functions
Trigonometric functions play a fundamental role in variable separation and integration processes. They often appear in physics and engineering problems that involve periodic or oscillating behavior.
In our solution, the condition \( y' \tan x = a + y \) involves the trigonometric function \( \tan x \). Upon separating variables and integrating, we make use of \( \int \cot x \, dx \), which is a common trigonometric integration leading us to \( \ln|\sin x| \).
When applying the initial condition at \( x = \pi/3 \), it helps to know that \( \sin(\pi/3) = \frac{\sqrt{3}}{2} \). These trigonometric identities simplify the application of conditions and the final solving of the differential equation for a specific solution, considering the periodic nature of these functions.

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Most popular questions from this chapter

Find all equilibria of the autonomous differential equation and construct the phase plot. (a) $$ y^{\prime}=y^{2}-2$$ (b)$$y^{\prime}=\frac{y-3}{y+9}, \quad y \geqslant 0$$ (c) $$y^{\prime}=y(3-y)\left(25-y^{2}\right)$$

(a) What can you say about a solution of the equation \(y^{\prime}=-y^{2}\) just by looking at the differential equation? (b) Verify that all members of the family \(y=1 /(x+C)\) are solutions of the equation in part (a). (c) Can you think of a solution of the differential equation \(y^{\prime}=-y^{2}\) that is not a member of the family in part (b)? (d) Find a solution of the initial-value problem $$y^{\prime}=-y^{2} \quad y(0)=0.5$$

Solve the equation \(e ^ { - y } y ^ { \prime } + \cos x = 0\) and graph several members of the family of solutions. How does the solution curve change as the constant \(C\) varies?

Suppose that the population dynamics of a species obeys a modified version of the logistic differential equation having the following form: \(\frac{d N}{d t}=r\left(1-\frac{N}{K}\right)^{2} N\) where \(r \neq 0\) and \(K>0\) (a) Show that \(\hat{N}=0\) and \(\hat{N}=K\) are equilibria. (b) For which values of \(r\) is the equilibrium \(\hat{N}=0\) unstable? (b) For which values of \(r\) is the equilibrium \(\hat{N}=0\) unstable? (c) Apply the local stability criterion to the equilibrium \(\hat{N}=K .\) What do you think your answer means about the stability of this equilibrium? (Note: This is an example in which the local stability criterion is inconclusive.) (d) Construct two phase plots, one for the case where \(r>0\) and the other for \(r<0,\) and determine the stability of \(\hat{N}=K\) in each case. Does the answer match your reasoning in part \((c) ?\)

Hooke's Law states that the force \(F\) exerted by a spring on a mass is proportional to the displacement from its resting position. From the figure we have \(F=-k p\) for some positive constant \(k,\) where \(p\) is position. Also, Newton's Second Law tells us that \(F=m a,\) where \(m\) is the mass of the object and \(a\) is its acceleration. Further, since we know that acceleration is the second derivative of position with respect to time, we obtain the differential equation \(m \frac{d^{2} p}{d t^{2}}=-k p\) This is a second-order differential equation because it involves the second derivative of the unknown function, \(p\) . (a) Define a new variable \(q=d p / d t\) . Show that the second-order differential equation can then be written a sthe following system of two first-order differential equations \(\frac{d p}{d t}=q \quad \frac{d q}{d t}=-\frac{k p}{m}\) (b) Construct the phase plane for the equations in part (a), including nullclines, the equilibrium, and the direction of movement. (c) What does the phase plane analysis from part (b) tell you about the position of the mass over time? (d) What does the phase plane analysis from part (b) tell you about the velocity of the mass over time?

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