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In Problems 9-16, solve the given differential equation. $$ \left(y^{2}+1\right) d x=y \sec ^{2} x d y $$

Short Answer

Expert verified
The solution is \( y - \frac{1}{y} = \frac{x}{2} + \frac{1}{4}\sin(2x) + C \).

Step by step solution

01

Identify the type of differential equation

The given equation is \((y^2 + 1) \, dx = y \sec^2 x \, dy\). This is not a separable equation in the form it is presented, but by rearranging, it can be transformed to a separable form.
02

Rearrange the equation

Rearrange the equation to isolate terms involving \(x\) on one side and terms involving \(y\) on the other side: \[ \frac{dy}{dx} = \frac{y \sec^2 x}{y^2 + 1} \] This is now a first-order differential equation that can be separated into differentials.
03

Separate the variables

Separate the variables by multiplying both sides by \(\frac{1}{y}\) and \(\cos^2 x\) to isolate each variable:\[ \frac{(y^2+1)}{y} \, dy = \cos^2 x \, dx \]
04

Integrate both sides

Integrate both sides separately:- For the left side: \( \int \left(1 + \frac{1}{y^2}\right) \, dy = y - \frac{1}{y} + C_1 \) - For the right side: \( \int \cos^2 x \, dx \) Apply the identity \(\cos^2 x = \frac{1 + \cos(2x)}{2}\): \[ \int \left(\frac{1}{2} + \frac{1}{2}\cos(2x)\right) \, dx = \frac{x}{2} + \frac{1}{4}\sin(2x) + C_2 \]
05

Combine the integrated results

Equate the results of the integration to find the solution:\[ y - \frac{1}{y} + C_1 = \frac{x}{2} + \frac{1}{4}\sin(2x) + C_2 \] Combine the constants: \( C = C_2 - C_1 \), giving:\[ y - \frac{1}{y} = \frac{x}{2} + \frac{1}{4}\sin(2x) + C \]

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Key Concepts

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

Separable Equations
Differential equations can often be complex and tricky to solve in their original form. However, a special type known as separable equations makes the task simpler.
These equations feature expressions where you can separate the variables into distinct groups, each involving only one of the variables with their respective differential. This separation allows us to integrate both sides independently.

Imagine this analogy: you have a recipe where all your dry ingredients (like flour and sugar) and wet ingredients (like milk and eggs) must be mixed separately before combining everything to finish the cake. In mathematics, it’s somewhat similar—variables are separated to process each independently before combining their results to find the solution.

Our original equation, though not initially separable, became one by rearranging it. This is an important skill—finding ways to transform equations to fit the separable structure.
First-order Differential Equation
A first-order differential equation involves the first derivative of the function. These equations typically appear in forms that allow for straightforward (or relatively so) mathematical techniques to solve.
In our case, the first-order differential equation arises after some algebra:
  • We transform the original differential equation by rearranging terms.
  • The aim is to isolate the \( \frac{dy}{dx} \) to identify the behavioral relationship between the derived function and its first derivative.
This mathematical process guarantees we understand the change of the system or function, which the equation depicts.
This step is crucial as it sets the stage for separation by making the variables ‘friendlier’ to isolate and integrate.
Integration Techniques
Once we have separated our first-order differential equation, integration techniques become the next focus. Each class of integrals can require different methods, depending on the function’s complexity.
For the left-hand side in our solution, we used basic integration laws: \( \int (1 + \frac{1}{y^2}) \, dy = y - \frac{1}{y} + C_1 \). The approach here involves recognizing the integral form, integrating constant terms, and handling polynomial-like functions.

For the right-hand side, however, we used trigonometric identities to ease the process: \[ \int \cos^2 x \, dx \]. By applying the identity \(\cos^2 x = \frac{1 + \cos(2x)}{2} \), the integration becomes manageable. After simplifying using this identity, the result \( \frac{x}{2} + \frac{1}{4}\sin(2x) + C_2 \) is achieved.

This flexibility in integration methods showcases the richness in approaching solutions: recognizing when to use standard methods or when a transformation is necessary.
Variable Separation
Variable separation is at the heart of solving a separable equation.
Once identified as a separable equation, the goal is straightforward: rearrange the terms to have all instances and differentials of one variable on each side of the equation. In our exercise, we multiplied each side to have \( \frac{(y^2+1)}{y} \, dy = \cos^2 x \, dx \), achieving a neat separation.
  • This approach allows functions involving \ x \ to stay independent from functions involving \ y \.
  • Thus, each side can then be dealt with independently through integration.
Variable separation simplifies a potentially large system by focusing on smaller, manageable parts.
This methodical breakdown is akin to decluttering a room for easier cleaning—clarifying distinct areas leads to a more organized and focused process for finding the solution.

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Most popular questions from this chapter

Rocket Motion Suppose a small single-stage rocket of total mass \(m(t)\) is launched vertically and that the rocket consumes its fuel at a constant rate. If the positive direction is upward and if we take air resistance to be linear, then a differential equation for its velocity \(v(t)\) is given by $$ \frac{d v}{d t}+\frac{k-\lambda}{m_{0}-\lambda t} v=-g+\frac{R}{m_{0}-\lambda t^{\prime}} $$ where \(k\) is the drag coefficient, \(\lambda\) is the rate at which fuel is consumed, \(R\) is the thrust of the rocket, \(m_{0}\) is the total mass of the rocket at \(t=0\), and \(g\) is the acceleration due to gravity. See Problem 21 in Exercises \(1.3\). (a) Find the velocity \(v(t)\) of the rocket if \(m_{0}=200 \mathrm{~kg}\), \(R=2000 \mathrm{~N}, \lambda=1 \mathrm{~kg} / \mathrm{s}, g=9.8 \mathrm{~m} / \mathrm{s}^{2}, k=3 \mathrm{~kg} / \mathrm{s}\), and \(v(0)=0\) (b) Use \(d s / d t=v\) and the result in part (a) to find the height \(s(t)\) of the rocket at time \(t\).

Find the general solution of the given differential equation. Give the largest interval over which the general solution is defined. Determine whether there are any transient terms in the general solution. $$ \cos ^{2} x \sin x \frac{d y}{d x}+\left(\cos ^{3} x\right) y=1 $$

Express the solution of the initial-value problem \(y^{\prime}-2 x y=1\), \(y(1)=1\), in terms of \(\operatorname{erf}(x)\)

The differential equation $$ \frac{d y}{d x}=\frac{-x+\sqrt{x^{2}+y^{2}}}{y} $$ describes the shape of a plane curve \(C\) that will reflect all incoming light beams to the same point and could be a model for the mirror of a reflecting telescope, a satellite antenna, or a solar collector. See Problem 29 in Exercises \(1.3\). There are several ways of solving this \(\mathrm{DE}\). (a) Verify that the differential equation is homogeneous (see Section 2.5). Show that the substitution \(y=u x\) yields $$ \frac{u d u}{\sqrt{1+u^{2}}\left(1-\sqrt{\left.1+u^{2}\right)}\right.}=\frac{d x}{x}. $$ Use a CAS (or another judicious substitution) to integrate the left-hand side of the equation. Show that the curve \(C\) must be a parabola with focus at the origin and is symmetric with respect to the \(x\) -axis. (b) Show that the first differential equation can also be solved by means of the substitution \(u=x^{2}+y^{2}\).

An electromotive force $$ E(t)=\left\\{\begin{array}{ll} 120, & 0 \leq t \leq 20 \\ 0, & t>20 \end{array}\right. $$ is applied to an \(L R\) -series circuit in which the inductance is 20 henries and the resistance is 2 ohms. Find the current \(i(t)\) if \(i(0)=0\).

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