Solve The Differential Equation: ${ \frac{dy}{dx} = \frac{y \cos X}{1 + 2y^2} }$A. { \ln |y| + Y^2 = \sin X + C$}$B. { \ln |y| - Y^2 = \sin X + C$}$C. { Y^2 = \sin X + C$}$D. None Of These.

by ADMIN 191 views

Introduction

Differential equations are a fundamental concept in mathematics, and solving them is a crucial skill for students and professionals alike. In this article, we will focus on solving a specific differential equation, which is given by:

dydx=ycos⁡x1+2y2\frac{dy}{dx} = \frac{y \cos x}{1 + 2y^2}

This equation is a first-order differential equation, and it involves a function yy that depends on the variable xx. Our goal is to find the general solution of this equation, which will give us the function yy in terms of xx.

Separating the Variables

To solve this differential equation, we will use the method of separation of variables. This method involves rearranging the equation so that all the terms involving yy are on one side, and all the terms involving xx are on the other side.

Let's start by multiplying both sides of the equation by the denominator, which is 1+2y21 + 2y^2. This gives us:

(1+2y2)dydx=ycos⁡x(1 + 2y^2) \frac{dy}{dx} = y \cos x

Now, we can rearrange the equation to get:

dydx=ycos⁡x1+2y2\frac{dy}{dx} = \frac{y \cos x}{1 + 2y^2}

Integrating the Equation

The next step is to integrate both sides of the equation with respect to xx. This will give us:

∫dydxdx=∫ycos⁡x1+2y2dx\int \frac{dy}{dx} dx = \int \frac{y \cos x}{1 + 2y^2} dx

Using the substitution method, we can rewrite the right-hand side of the equation as:

∫ycos⁡x1+2y2dx=∫ycos⁡x1+2y2⋅1ydy\int \frac{y \cos x}{1 + 2y^2} dx = \int \frac{y \cos x}{1 + 2y^2} \cdot \frac{1}{y} dy

This simplifies to:

∫dydxdx=∫cos⁡x1+2y2dy\int \frac{dy}{dx} dx = \int \frac{\cos x}{1 + 2y^2} dy

Evaluating the Integral

Now, we need to evaluate the integral on the right-hand side. This is a bit tricky, but we can use the substitution method again. Let's substitute u=1+2y2u = 1 + 2y^2. Then, we have:

du=4ydydu = 4y dy

Rearranging this equation, we get:

dy=du4ydy = \frac{du}{4y}

Substituting this into the integral, we get:

∫cos⁡x1+2y2dy=∫cos⁡xu⋅du4y\int \frac{\cos x}{1 + 2y^2} dy = \int \frac{\cos x}{u} \cdot \frac{du}{4y}

This simplifies to:

∫cos⁡x1+2y2dy=14∫cos⁡xudu\int \frac{\cos x}{1 + 2y^2} dy = \frac{1}{4} \int \frac{\cos x}{u} du

Solving for y

Now, we can evaluate the integral on the right-hand side. This gives us:

14∫cos⁡xudu=14∫cos⁡x1+2y2du\frac{1}{4} \int \frac{\cos x}{u} du = \frac{1}{4} \int \frac{\cos x}{1 + 2y^2} du

Using the substitution method again, we can rewrite this as:

14∫cos⁡x1+2y2du=14∫cos⁡x1+2y2⋅12ydy\frac{1}{4} \int \frac{\cos x}{1 + 2y^2} du = \frac{1}{4} \int \frac{\cos x}{1 + 2y^2} \cdot \frac{1}{2y} dy

This simplifies to:

14∫cos⁡x1+2y2du=18y∫cos⁡xdx\frac{1}{4} \int \frac{\cos x}{1 + 2y^2} du = \frac{1}{8y} \int \cos x dx

Evaluating the Integral

Now, we need to evaluate the integral on the right-hand side. This is a standard integral, and it gives us:

18y∫cos⁡xdx=18ysin⁡x+C\frac{1}{8y} \int \cos x dx = \frac{1}{8y} \sin x + C

Simplifying the Solution

Now, we can simplify the solution by multiplying both sides by 8y8y. This gives us:

sin⁡x+C=8y2\sin x + C = 8y^2

Rearranging the Solution

Finally, we can rearrange the solution to get:

y2=18sin⁡x+C8y^2 = \frac{1}{8} \sin x + \frac{C}{8}

Conclusion

In this article, we have solved the differential equation:

dydx=ycos⁡x1+2y2\frac{dy}{dx} = \frac{y \cos x}{1 + 2y^2}

Using the method of separation of variables, we have found the general solution of this equation, which is:

y2=18sin⁡x+C8y^2 = \frac{1}{8} \sin x + \frac{C}{8}

This solution is a quadratic function of yy, and it involves the sine function of xx. We can see that the solution involves a constant term CC, which is a characteristic of the general solution of a differential equation.

Answer

The correct answer is:

C. y2=sin⁡x+Cy^2 = \sin x + C

This is the general solution of the differential equation, and it involves the sine function of xx.

Introduction

In our previous article, we solved the differential equation:

dydx=ycos⁡x1+2y2\frac{dy}{dx} = \frac{y \cos x}{1 + 2y^2}

Using the method of separation of variables, we found the general solution of this equation, which is:

y2=18sin⁡x+C8y^2 = \frac{1}{8} \sin x + \frac{C}{8}

In this article, we will answer some common questions that students and professionals may have about solving differential equations.

Q: What is the method of separation of variables?

A: The method of separation of variables is a technique used to solve differential equations. It involves rearranging the equation so that all the terms involving one variable are on one side, and all the terms involving the other variable are on the other side.

Q: How do I know when to use the method of separation of variables?

A: You can use the method of separation of variables when the differential equation can be written in the form:

dydx=f(x)g(y)\frac{dy}{dx} = f(x) g(y)

where f(x)f(x) and g(y)g(y) are functions of xx and yy respectively.

Q: What is the general solution of a differential equation?

A: The general solution of a differential equation is a solution that involves an arbitrary constant. In the case of the differential equation we solved earlier, the general solution is:

y2=18sin⁡x+C8y^2 = \frac{1}{8} \sin x + \frac{C}{8}

Q: How do I find the particular solution of a differential equation?

A: To find the particular solution of a differential equation, you need to know the initial conditions. The initial conditions are the values of the variables at a specific point in time. Once you have the initial conditions, you can substitute them into the general solution to find the particular solution.

Q: What is the difference between a general solution and a particular solution?

A: The general solution of a differential equation is a solution that involves an arbitrary constant, while the particular solution is a solution that is specific to a particular set of initial conditions.

Q: Can I use the method of separation of variables to solve all types of differential equations?

A: No, the method of separation of variables is only applicable to certain types of differential equations. It is not applicable to differential equations that involve higher-order derivatives or nonlinear terms.

Q: How do I know if a differential equation can be solved using the method of separation of variables?

A: You can try to rearrange the equation so that all the terms involving one variable are on one side, and all the terms involving the other variable are on the other side. If you can do this, then the method of separation of variables may be applicable.

Q: What are some common mistakes to avoid when solving differential equations?

A: Some common mistakes to avoid when solving differential equations include:

  • Not checking the initial conditions
  • Not using the correct method for solving the differential equation
  • Not checking the solution for consistency
  • Not using the correct notation

Conclusion

In this article, we have answered some common questions that students and professionals may have about solving differential equations. We have also provided some tips and tricks for solving differential equations using the method of separation of variables.

Final Answer

The final answer is:

C. y2=sin⁡x+Cy^2 = \sin x + C

This is the general solution of the differential equation, and it involves the sine function of xx.