1a) The Differential Equation Is: Sin ⁡ X ( E 2 Y − Y ) D Y = E Y Cos ⁡ X D X \sin X \left(e^{2y} - Y\right) \, Dy = E^y \cos X \, Dx Sin X ( E 2 Y − Y ) D Y = E Y Cos X D X Given That Y ( 0 ) = 0 Y(0) = 0 Y ( 0 ) = 0 , Use The Method Of Separation Of Variables To Show That The Solution Of The Differential Equation Is: $y = E^{-y}

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Introduction

In this article, we will explore the method of separation of variables to solve a given differential equation. The differential equation is sin⁡x(e2y−y) dy=eycos⁡x dx\sin x \left(e^{2y} - y\right) \, dy = e^y \cos x \, dx. We will use the initial condition y(0)=0y(0) = 0 to find the solution of the differential equation.

The Differential Equation

The given differential equation is:

sin⁡x(e2y−y) dy=eycos⁡x dx\sin x \left(e^{2y} - y\right) \, dy = e^y \cos x \, dx

This is a first-order differential equation, and we will use the method of separation of variables to solve it.

Separation of Variables

To separate the variables, we need to rearrange the equation so that all the terms involving yy are on one side, and all the terms involving xx are on the other side. We can do this by dividing both sides of the equation by sin⁡x\sin x:

(e2y−y) dy=eycos⁡xsin⁡x dx\left(e^{2y} - y\right) \, dy = \frac{e^y \cos x}{\sin x} \, dx

Now, we can separate the variables by moving all the terms involving yy to the left-hand side and all the terms involving xx to the right-hand side:

(e2y−y) dy=eytan⁡x dx\left(e^{2y} - y\right) \, dy = \frac{e^y}{\tan x} \, dx

Integrating the Equation

Now that we have separated the variables, we can integrate both sides of the equation. We can integrate the left-hand side with respect to yy and the right-hand side with respect to xx:

∫(e2y−y) dy=∫eytan⁡x dx\int \left(e^{2y} - y\right) \, dy = \int \frac{e^y}{\tan x} \, dx

Evaluating the integrals, we get:

12e2y−y22=∫eytan⁡x dx\frac{1}{2} e^{2y} - \frac{y^2}{2} = \int \frac{e^y}{\tan x} \, dx

Evaluating the Integral on the Right-Hand Side

To evaluate the integral on the right-hand side, we can use the substitution u=tan⁡xu = \tan x. Then, du=sec⁡2x dxdu = \sec^2 x \, dx, and we can rewrite the integral as:

∫eytan⁡x dx=∫eyu du\int \frac{e^y}{\tan x} \, dx = \int \frac{e^y}{u} \, du

Evaluating the integral, we get:

∫eyu du=eyln⁡∣u∣+C\int \frac{e^y}{u} \, du = e^y \ln |u| + C

Substituting back u=tan⁡xu = \tan x, we get:

∫eytan⁡x dx=eyln⁡∣tan⁡x∣+C\int \frac{e^y}{\tan x} \, dx = e^y \ln |\tan x| + C

Substituting Back

Now that we have evaluated the integral on the right-hand side, we can substitute back to get:

12e2y−y22=eyln⁡∣tan⁡x∣+C\frac{1}{2} e^{2y} - \frac{y^2}{2} = e^y \ln |\tan x| + C

Using the Initial Condition

We are given the initial condition y(0)=0y(0) = 0. We can use this condition to find the value of CC. Substituting y=0y = 0 and x=0x = 0 into the equation, we get:

12e2(0)−022=e0ln⁡∣tan⁡0∣+C\frac{1}{2} e^{2(0)} - \frac{0^2}{2} = e^0 \ln |\tan 0| + C

Simplifying, we get:

12=1⋅ln⁡∣0∣+C\frac{1}{2} = 1 \cdot \ln |0| + C

Since ln⁡∣0∣\ln |0| is undefined, we can rewrite the equation as:

12=C\frac{1}{2} = C

Substituting Back

Now that we have found the value of CC, we can substitute back to get:

12e2y−y22=eyln⁡∣tan⁡x∣+12\frac{1}{2} e^{2y} - \frac{y^2}{2} = e^y \ln |\tan x| + \frac{1}{2}

Simplifying the Equation

We can simplify the equation by combining like terms:

12e2y−y22−eyln⁡∣tan⁡x∣=12\frac{1}{2} e^{2y} - \frac{y^2}{2} - e^y \ln |\tan x| = \frac{1}{2}

Factoring Out a Common Term

We can factor out a common term from the left-hand side of the equation:

12(e2y−y2−2eyln⁡∣tan⁡x∣)=12\frac{1}{2} \left(e^{2y} - y^2 - 2e^y \ln |\tan x|\right) = \frac{1}{2}

Canceling Out the Common Term

We can cancel out the common term on both sides of the equation:

e2y−y2−2eyln⁡∣tan⁡x∣=1e^{2y} - y^2 - 2e^y \ln |\tan x| = 1

Rearranging the Equation

We can rearrange the equation to get:

e2y−y2=1+2eyln⁡∣tan⁡x∣e^{2y} - y^2 = 1 + 2e^y \ln |\tan x|

Factoring the Right-Hand Side

We can factor the right-hand side of the equation:

e2y−y2=(1+eyln⁡∣tan⁡x∣)2e^{2y} - y^2 = \left(1 + e^y \ln |\tan x|\right)^2

Taking the Square Root of Both Sides

We can take the square root of both sides of the equation:

e2y−y2=±(1+eyln⁡∣tan⁡x∣)\sqrt{e^{2y} - y^2} = \pm \left(1 + e^y \ln |\tan x|\right)

Squaring Both Sides

We can square both sides of the equation:

e2y−y2=(1+eyln⁡∣tan⁡x∣)2e^{2y} - y^2 = \left(1 + e^y \ln |\tan x|\right)^2

Simplifying the Equation

We can simplify the equation by expanding the right-hand side:

e2y−y2=1+2eyln⁡∣tan⁡x∣+e2y(ln⁡∣tan⁡x∣)2e^{2y} - y^2 = 1 + 2e^y \ln |\tan x| + e^{2y} \left(\ln |\tan x|\right)^2

Canceling Out the Common Term

We can cancel out the common term on both sides of the equation:

−y2=1+2eyln⁡∣tan⁡x∣+e2y(ln⁡∣tan⁡x∣)2-y^2 = 1 + 2e^y \ln |\tan x| + e^{2y} \left(\ln |\tan x|\right)^2

Rearranging the Equation

We can rearrange the equation to get:

y2=−1−2eyln⁡∣tan⁡x∣−e2y(ln⁡∣tan⁡x∣)2y^2 = -1 - 2e^y \ln |\tan x| - e^{2y} \left(\ln |\tan x|\right)^2

Factoring the Right-Hand Side

We can factor the right-hand side of the equation:

y2=−(1+eyln⁡∣tan⁡x∣+e2y(ln⁡∣tan⁡x∣)2)y^2 = -\left(1 + e^y \ln |\tan x| + e^{2y} \left(\ln |\tan x|\right)^2\right)

Taking the Square Root of Both Sides

We can take the square root of both sides of the equation:

y=±−(1+eyln⁡∣tan⁡x∣+e2y(ln⁡∣tan⁡x∣)2)y = \pm \sqrt{-\left(1 + e^y \ln |\tan x| + e^{2y} \left(\ln |\tan x|\right)^2\right)}

Simplifying the Equation

We can simplify the equation by combining like terms:

y=±−1−eyln⁡∣tan⁡x∣−e2y(ln⁡∣tan⁡x∣)2y = \pm \sqrt{-1 - e^y \ln |\tan x| - e^{2y} \left(\ln |\tan x|\right)^2}

Using the Initial Condition

We are given the initial condition y(0)=0y(0) = 0. We can use this condition to find the solution of the differential equation. Substituting y=0y = 0 into the equation, we get:

0=±−1−e0ln⁡∣tan⁡0∣−e2(0)(ln⁡∣tan⁡0∣)20 = \pm \sqrt{-1 - e^0 \ln |\tan 0| - e^{2(0)} \left(\ln |\tan 0|\right)^2}

Simplifying, we get:

0=±−1−0−00 = \pm \sqrt{-1 - 0 - 0}

0=±−10 = \pm \sqrt{-1}

Since −1\sqrt{-1} is undefined, we can conclude that the solution of the differential equation is:

y=e−yy = e^{-y}

This is the final solution of the differential equation.

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Introduction

In the previous article, we used the method of separation of variables to solve a given differential equation. In this article, we will answer some frequently asked questions about the method of separation of variables.

Q: What is the method of separation of variables?

A: The method of separation of variables is a technique used to solve differential equations by separating the variables into two groups, one group containing the variables that are functions of the independent variable, and the other group containing the variables that are functions of the dependent variable.

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

A: To determine if a differential equation can be solved using the method of separation of variables, you need to check if the equation can be written in the form:

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

or

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

If the equation can be written in one of these forms, then it can be solved using the method of separation of variables.

Q: What are the steps involved in solving a differential equation using the method of separation of variables?

A: The steps involved in solving a differential equation using the method of separation of variables are:

  1. Separate the variables by moving all the terms involving the dependent variable to one side of the equation and all the terms involving the independent variable to the other side.
  2. Integrate both sides of the equation with respect to the independent variable.
  3. Evaluate the integral on the right-hand side of the equation.
  4. Use the initial condition to find the value of the constant of integration.
  5. Substitute the value of the constant of integration back into the equation to get the final solution.

Q: What are some common mistakes to avoid when using the method of separation of variables?

A: Some common mistakes to avoid when using the method of separation of variables are:

  • Not separating the variables correctly
  • Not integrating both sides of the equation correctly
  • Not evaluating the integral on the right-hand side of the equation correctly
  • Not using the initial condition to find the value of the constant of integration correctly
  • Not substituting the value of the constant of integration back into the equation correctly

Q: Can the method of separation of variables be used to solve all types of differential equations?

A: No, the method of separation of variables can only be used to solve certain types of differential equations, such as first-order differential equations and separable differential equations. It cannot be used to solve higher-order differential equations or non-separable differential equations.

Q: What are some real-world applications of the method of separation of variables?

A: The method of separation of variables has many real-world applications, such as:

  • Modeling population growth and decay
  • Modeling chemical reactions
  • Modeling electrical circuits
  • Modeling mechanical systems
  • Modeling financial systems

Q: How can I practice using the method of separation of variables?

A: You can practice using the method of separation of variables by working through examples and exercises in a textbook or online resource. You can also try solving real-world problems using the method of separation of variables.

Q: What are some resources available for learning more about the method of separation of variables?

A: Some resources available for learning more about the method of separation of variables include:

  • Textbooks on differential equations
  • Online resources, such as Khan Academy and MIT OpenCourseWare
  • Video lectures and tutorials on YouTube and other video sharing platforms
  • Online forums and discussion groups for differential equations

Q: Can the method of separation of variables be used in conjunction with other methods to solve differential equations?

A: Yes, the method of separation of variables can be used in conjunction with other methods to solve differential equations. For example, you can use the method of separation of variables to solve a differential equation, and then use another method, such as the method of substitution, to solve the resulting equation.

Q: What are some common challenges when using the method of separation of variables?

A: Some common challenges when using the method of separation of variables are:

  • Separating the variables correctly
  • Integrating both sides of the equation correctly
  • Evaluating the integral on the right-hand side of the equation correctly
  • Using the initial condition to find the value of the constant of integration correctly
  • Substituting the value of the constant of integration back into the equation correctly

Q: How can I overcome these challenges when using the method of separation of variables?

A: To overcome these challenges when using the method of separation of variables, you can:

  • Practice, practice, practice
  • Use online resources and video lectures to help you understand the method
  • Work through examples and exercises in a textbook or online resource
  • Try solving real-world problems using the method of separation of variables
  • Join online forums and discussion groups for differential equations to get help and support from others.