/*! 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 52 \(\frac{d y}{d \theta}=\frac{\si... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 52

\(\frac{d y}{d \theta}=\frac{\sin \theta}{\cos y+1}, y(0)=0\)

Short Answer

Expert verified
The solution is \ y=-\frac{1}{text{cos} \frac1}{1\text cos}}\(y= solved as1$\. dy\frac-id)}}\

Step by step solution

01

- Identify the type of differential equation

Recognize that the given equation \(\frac{d y}{d \theta}=\frac{\text{sin} \theta}{\text{cos} y+1}\) is a first-order ordinary differential equation.
02

- Separate the variables

Rewrite the equation to separate the variables \(\theta\) and \(y\): \(\frac{d y}{\text{cos} y + 1} = \text{sin} \theta \, d\theta\).
03

- Integrate both sides

Integrate both sides of the separated equation to find \(y\). For the left side: \(\begin{align*}\frac{1}{\text{cos} y + 1} \frac{dy}{dx} &= \frac{\text{sin} \theta}{\text{cos} y + 1}\frac{d y}{dy}\text{\Then, integrate:}\text{\right(k\text{\=-1}+\frac{1}{\text}=(\frac{\text e}{\frac{dy=dx\theta)=1x(a(val\theta?ld\right=by\theta}})}}\)\frac {1+ y}}}}=+{u(1)by=x}\text y =text

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Key Concepts

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

Separation of Variables
Separation of Variables is a method used to solve first-order ordinary differential equations. This technique involves rearranging the given equation so that each variable and its differential are on opposite sides of the equation. This method works best when both sides can be integrated independently. In our example, \(\frac{d y}{d \theta}=\frac{\text{sin} \theta}{\text{cos} y+1}\), the goal is to get all the y terms (including dy) on one side and all the 胃 terms (including d胃) on the other. We start by rewriting the equation: \(\frac{d y}{\text{cos} y + 1} = \text{sin} \theta \, d \theta\). Now, the variables are separated, allowing integration to be the next step.
Integration
Once variables are separated, the next step is integration. This involves finding the antiderivatives of both sides. For the left side, \(\frac{1}{\text{cos} y+1}\frac{d y}{d y}\), we need to integrate with respect to y. Similarly, for the right side, \(\text{sin} \theta \, d\theta\), we integrate with respect to 胃. Suppose: Let u = \(\text{cos} y + 1\). Integrate both sides to get: \(\text{integral of}\frac{d y}{\text{cos} y + 1} = \text{integral of}\text{sin} \theta \, d \theta\). Solving these integrals will help us find an explicit relation between y and 胃.
Initial Conditions
Initial Conditions are essential for finding the specific solution to a differential equation. They provide a starting point, usually given in the form of \((\theta_0, y_0)\). In the example, we have the initial condition \(y(0) = 0\). Once you have integrated and found the general solution, you apply the initial condition to solve for any constants \(C\). This gives you the specific solution that satisfies both the differential equation and the initial condition. Using the initial condition \(y(0) = 0\), we substitute \( \theta = 0\) and \(y = 0\) into the solved equation to find the particular constant.

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Most popular questions from this chapter

Find the family of curves tangent to the force field $$ \mathbf{F}(x, y)=\underbrace{\frac{x^{2}-y^{2}}{\sqrt{x^{2}+y^{2}}}}_{d x / d t} \mathbf{i}-\underbrace{\frac{2 x y}{\sqrt{x^{2}+y^{2}}}}_{d y / d t} \mathbf{j} . $$

\(y^{4} d t+\left(t^{4}-t y^{3}\right) d y=0, y(1)=2\)

\(x d y / d x+y=x e^{x}\)

\(t y^{\prime}+y=t\)

If \(M(t, y) d t+N(t, y) d y=0\) is a homogeneous equation, show that the change of variables \(x=r \cos \theta\) and \(y=r \sin \theta\) transform the homogeneous equation into a separable equation.
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