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Solve the differential equation by separation of variables. Where reasonable, express the family of solutions as explicit functions of \(x\). $$y-\frac{d y}{d x} \sec x=0$$

Short Answer

Expert verified
The solution is \( y = C e^{\sin x} \).

Step by step solution

01

Rearrange the Differential Equation

Start by reorganizing the differential equation:\[ y - \frac{dy}{dx} \sec x = 0 \]This can be rearranged as:\[ \frac{dy}{dx} \sec x = y \]
02

Separate Variables

The goal is to separate variables so that all terms involving \(y\) are on one side and all terms involving \(x\) are on the other side. Start by rewriting the equation:\[ \frac{dy}{dx} = y \cos x \]Now separate the variables by dividing both sides by \(y\) and multiplying both sides by \(dx\):\[ \frac{1}{y} dy = \cos x \ dx \]
03

Integrate Both Sides

Integrate both sides to solve for \(y\). The integral of \(\frac{1}{y} dy\) is \(\ln |y|\), and the integral of \(\cos x \ dx\) is \(\sin x + C_1\), where \(C_1\) is the constant of integration:\[ \int \frac{1}{y} dy = \int \cos x \ dx \]\[ \ln |y| = \sin x + C_1 \]
04

Solve For y

To express \(y\) as an explicit function of \(x\), exponentiate both sides to eliminate the natural logarithm:\[ e^{\ln |y|} = e^{\sin x + C_1} \]\[ |y| = e^{C_1} \cdot e^{\sin x} \]Let \(C = \pm e^{C_1}\), which is a new constant that accounts for the absolute value:\[ y = C e^{\sin x} \]

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Key Concepts

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

Separation of Variables
The method called "separation of variables" is a powerful technique for solving differential equations. This method is especially useful when an equation can be written in a form that allows us to separate the variables, meaning we can isolate terms involving "x" on one side of the equation and terms involving "y" on the other side.
In differential equations, separation of variables involves a few key steps:
  • Rearrange the equation so that each variable and its derivative are on opposite sides.
  • Integrate each side separately to find a solution that involves a constant of integration.
In our exercise, this is seen through the step-by-step solution where the equation is rearranged to \( \frac{1}{y} \, dy = \cos x \, dx \), allowing us to then integrate both sides independently. This separation is vital because it allows us to solve the parts one at a time, which is often simpler than trying to solve the entire equation in its initial form.
Integrating Factor
Although we did not use an integrating factor in this problem directly, it is worth knowing what it is since integrating factors are common in solving linear differential equations of the form \( \frac{dy}{dx} + P(x)y = Q(x) \). In these types of equations, finding an integrating factor can simplify the process of finding a solution. An integrating factor, \( \mu(x) \), is a function that we multiply through the entire differential equation to make the left-hand side an exact derivative.
  • Calculate the integrating factor, typically using \( e^{\int P(x) \, dx} \).
  • Multiply every term in the equation by this integrating factor.
  • Use the fact that the left side now represents \( \frac{d}{dx}(\mu(x)y) = \mu(x)Q(x) \) to integrate and solve for \(y\).
Though it wasn't applicable to our given separable equation, understanding integrating factors equips us for solving more complex linear equations efficiently.
Explicit Functions
Expressing solutions as explicit functions is a common goal in solving differential equations, as explicit equations offer direct mathematical descriptions of relationships. An explicit function denotes "y" directly in terms of "x," without needing further manipulation.
In this exercise, we aimed to achieve such a function: starting with \( \ln |y| = \sin x + C_1 \), exponentiating both sides gave us an expression for \(y\) that is explicit: \( y = Ce^{\sin x} \).
  • Transform implicit solutions (involving logarithms or other nested expressions) into explicit forms for clarity.
  • Explicit forms are generally easier to analyze and graph, aiding in studying the behavior of solutions.
By converting logarithmic, implicit expressions to the form \(y = f(x)\), it becomes straightforward to understand how "y" changes as "x" varies, making these expressions vital in both theoretical and practical contexts.

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Most popular questions from this chapter

A rocket, fired upward from rest at time \(t=0,\) has an initial mass of \(m_{0}\) (including its fuel). Assuming that the fuel is consumed at a constant rate \(k\), the mass \(m\) of the rocket, while fuel is being burned, will be given by \(m=m_{0}-k t\) It can be shown that if air resistance is neglected and the fuel gases are expelled at a constant speed \(c\) relative to the rocket, then the velocity \(v\) of the rocket will satisfy the equation $$m \frac{d v}{d t}=c k-m g$$ where \(g\) is the acceleration due to gravity. (a) Find \(v(t)\) keeping in mind that the mass \(m\) is a function of \(t.\) (b) Suppose that the fuel accounts for \(80 \%\) of the initial mass of the rocket and that all of the fuel is consumed in 100 s. Find the velocity of the rocket in meters per second at the instant the fuel is exhausted. \([\) Note: Take \(\left.g=9.8 \mathrm{m} / \mathrm{s}^{2} \text { and } c=2500 \mathrm{m} / \mathrm{s} .\right]\)

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