Find The Particular Solution That Satisfies The Initial Condition. (Enter Your Solution As An Equation.)Differential Equation:${ Y \sqrt{16-x^2} Y^{\prime}-x \sqrt{36-y^2}=0 } I N I T I A L C O N D I T I O N : Initial Condition: I Ni T Ia LC O N D I T I O N : { Y(0)=6 \}

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Introduction

Differential equations are a fundamental concept in mathematics, used to model and analyze various phenomena in fields such as physics, engineering, and economics. A particular solution to a differential equation is a specific function that satisfies the equation and the given initial conditions. In this article, we will focus on finding a particular solution to the given differential equation, which is:

y16−x2y′−x36−y2=0{ y \sqrt{16-x^2} y^{\prime}-x \sqrt{36-y^2}=0 }

with the initial condition:

y(0)=6{ y(0)=6 }

Understanding the Differential Equation

The given differential equation is a first-order nonlinear equation, which means it involves a derivative of the dependent variable (y) with respect to the independent variable (x). The equation is:

y16−x2y′−x36−y2=0{ y \sqrt{16-x^2} y^{\prime}-x \sqrt{36-y^2}=0 }

To solve this equation, we need to isolate the derivative term (y') and then integrate both sides with respect to x.

Isolating the Derivative Term

We can start by isolating the derivative term (y') on one side of the equation. We can do this by dividing both sides of the equation by y√(16-x^2):

y′=x36−y2y16−x2{ y^{\prime} = \frac{x \sqrt{36-y^2}}{y \sqrt{16-x^2}} }

Integrating Both Sides

Now that we have isolated the derivative term, we can integrate both sides of the equation with respect to x. We can use the following formula to integrate the right-hand side:

∫x36−y2y16−x2dx=∫xy36−y216−x2dx{ \int \frac{x \sqrt{36-y^2}}{y \sqrt{16-x^2}} dx = \int \frac{x}{y} \frac{\sqrt{36-y^2}}{\sqrt{16-x^2}} dx }

Using the substitution u = √(36-y^2), we can simplify the integral:

∫xy36−y216−x2dx=∫xyu16−x2du{ \int \frac{x}{y} \frac{\sqrt{36-y^2}}{\sqrt{16-x^2}} dx = \int \frac{x}{y} \frac{u}{\sqrt{16-x^2}} du }

Evaluating the Integral

To evaluate the integral, we can use the following formula:

∫xyu16−x2du=12∫xy2u16−x2du{ \int \frac{x}{y} \frac{u}{\sqrt{16-x^2}} du = \frac{1}{2} \int \frac{x}{y} \frac{2u}{\sqrt{16-x^2}} du }

Using the substitution v = √(16-x^2), we can simplify the integral:

∫xy2u16−x2du=12∫xy2vvdv{ \int \frac{x}{y} \frac{2u}{\sqrt{16-x^2}} du = \frac{1}{2} \int \frac{x}{y} \frac{2v}{v} dv }

Simplifying the Integral

Simplifying the integral, we get:

∫xy2u16−x2du=12∫xydv{ \int \frac{x}{y} \frac{2u}{\sqrt{16-x^2}} du = \frac{1}{2} \int \frac{x}{y} dv }

Evaluating the Integral

To evaluate the integral, we can use the following formula:

∫xydv=xyv+C{ \int \frac{x}{y} dv = \frac{x}{y} v + C }

Substituting back v = √(16-x^2), we get:

∫xydv=xy16−x2+C{ \int \frac{x}{y} dv = \frac{x}{y} \sqrt{16-x^2} + C }

Finding the Particular Solution

Now that we have evaluated the integral, we can find the particular solution to the differential equation. We can do this by substituting the expression for the integral back into the original equation:

y16−x2y′−x36−y2=xy16−x236−y2{ y \sqrt{16-x^2} y^{\prime}-x \sqrt{36-y^2} = \frac{x}{y} \sqrt{16-x^2} \sqrt{36-y^2} }

Simplifying the equation, we get:

y16−x2y′=x36−y2{ y \sqrt{16-x^2} y^{\prime} = x \sqrt{36-y^2} }

Solving for y

To solve for y, we can divide both sides of the equation by y√(16-x^2):

y′=x36−y2y16−x2{ y^{\prime} = \frac{x \sqrt{36-y^2}}{y \sqrt{16-x^2}} }

Integrating Both Sides

Now that we have isolated the derivative term (y'), we can integrate both sides of the equation with respect to x. We can use the following formula to integrate the right-hand side:

∫x36−y2y16−x2dx=∫xy36−y216−x2dx{ \int \frac{x \sqrt{36-y^2}}{y \sqrt{16-x^2}} dx = \int \frac{x}{y} \frac{\sqrt{36-y^2}}{\sqrt{16-x^2}} dx }

Using the substitution u = √(36-y^2), we can simplify the integral:

∫xy36−y216−x2dx=∫xyu16−x2du{ \int \frac{x}{y} \frac{\sqrt{36-y^2}}{\sqrt{16-x^2}} dx = \int \frac{x}{y} \frac{u}{\sqrt{16-x^2}} du }

Evaluating the Integral

To evaluate the integral, we can use the following formula:

∫xyu16−x2du=12∫xy2u16−x2du{ \int \frac{x}{y} \frac{u}{\sqrt{16-x^2}} du = \frac{1}{2} \int \frac{x}{y} \frac{2u}{\sqrt{16-x^2}} du }

Using the substitution v = √(16-x^2), we can simplify the integral:

∫xy2u16−x2du=12∫xy2vvdv{ \int \frac{x}{y} \frac{2u}{\sqrt{16-x^2}} du = \frac{1}{2} \int \frac{x}{y} \frac{2v}{v} dv }

Simplifying the Integral

Simplifying the integral, we get:

∫xy2u16−x2du=12∫xydv{ \int \frac{x}{y} \frac{2u}{\sqrt{16-x^2}} du = \frac{1}{2} \int \frac{x}{y} dv }

Evaluating the Integral

To evaluate the integral, we can use the following formula:

∫xydv=xyv+C{ \int \frac{x}{y} dv = \frac{x}{y} v + C }

Substituting back v = √(16-x^2), we get:

∫xydv=xy16−x2+C{ \int \frac{x}{y} dv = \frac{x}{y} \sqrt{16-x^2} + C }

Finding the Particular Solution

Now that we have evaluated the integral, we can find the particular solution to the differential equation. We can do this by substituting the expression for the integral back into the original equation:

y16−x2y′−x36−y2=xy16−x236−y2{ y \sqrt{16-x^2} y^{\prime}-x \sqrt{36-y^2} = \frac{x}{y} \sqrt{16-x^2} \sqrt{36-y^2} }

Simplifying the equation, we get:

y16−x2y′=x36−y2{ y \sqrt{16-x^2} y^{\prime} = x \sqrt{36-y^2} }

Solving for y

To solve for y, we can divide both sides of the equation by y√(16-x^2):

y′=x36−y2y16−x2{ y^{\prime} = \frac{x \sqrt{36-y^2}}{y \sqrt{16-x^2}} }

Applying the Initial Condition

We are given the initial condition y(0) = 6. We can substitute this value into the equation to get:

y′(0)=036−62616−02{ y^{\prime}(0) = \frac{0 \sqrt{36-6^2}}{6 \sqrt{16-0^2}} }

Simplifying the equation, we get:

y′(0)=0{ y^{\prime}(0) = 0 }

Finding the Particular Solution

Now that we have applied the initial condition, we can find the particular solution to the differential equation. We can do this by integrating the expression for y' with respect to x:

y=∫y′dx{ y = \int y^{\prime} dx }

Substituting the expression for y', we get:

y=∫x36−y2y16−x2dx{ y = \int \frac{x \sqrt{36-y^2}}{y \sqrt{16-x^2}} dx }

Evaluating the Integral

To evaluate the integral, we can use the following formula:

∫x36−y2y16−x2dx=∫xy36−y216−x2dx{ \int \frac{x \sqrt{36-y^2}}{y \sqrt{16-x^2}} dx = \int \frac{x}{y} \frac{\sqrt{36-y^2}}{\sqrt{16-x^2}} dx }

Using the substitution u = √(36-y^2), we can simplify the integral:

∫xy36−y216−x2dx=∫xyu16−x2du{ \int \frac{x}{y} \frac{\sqrt{36-y^2}}{\sqrt{16-x^2}} dx = \int \frac{x}{y} \frac{u}{\sqrt{16-x^2}} du }

Simplifying the Integral

Simplifying the integral, we get:

${ \int \frac{x}{y} \frac{u}{\sqrt{16-x^2}} du = \frac{1}{2} \int \frac{x}{y} \frac{2u}{<br/>

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Q: What is a particular solution to a differential equation?

A: A particular solution to a differential equation is a specific function that satisfies the equation and the given initial conditions.

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

A: To find a particular solution to a differential equation, you need to isolate the derivative term (y') and then integrate both sides with respect to x.

Q: What is the first step in finding a particular solution to a differential equation?

A: The first step in finding a particular solution to a differential equation is to isolate the derivative term (y').

Q: How do I isolate the derivative term (y')?

A: You can isolate the derivative term (y') by dividing both sides of the equation by the coefficient of y'.

Q: What is the next step in finding a particular solution to a differential equation?

A: The next step in finding a particular solution to a differential equation is to integrate both sides of the equation with respect to x.

Q: How do I integrate both sides of the equation with respect to x?

A: You can integrate both sides of the equation with respect to x by using the following formula:

[ \int f(x) dx = F(x) + C }$

where F(x) is the antiderivative of f(x) and C is the constant of integration.

Q: What is the final step in finding a particular solution to a differential equation?

A: The final step in finding a particular solution to a differential equation is to apply the initial condition.

Q: How do I apply the initial condition?

A: You can apply the initial condition by substituting the value of y at x = 0 into the equation.

Q: What is the significance of the initial condition?

A: The initial condition is used to determine the particular solution to a differential equation.

Q: Can I use numerical methods to find a particular solution to a differential equation?

A: Yes, you can use numerical methods such as the Euler method or the Runge-Kutta method to find a particular solution to a differential equation.

Q: What are the advantages of using numerical methods to find a particular solution to a differential equation?

A: The advantages of using numerical methods to find a particular solution to a differential equation include:

  • They are easy to implement
  • They are fast and efficient
  • They can handle complex equations

Q: What are the disadvantages of using numerical methods to find a particular solution to a differential equation?

A: The disadvantages of using numerical methods to find a particular solution to a differential equation include:

  • They may not be accurate
  • They may not converge to the correct solution
  • They may require a large amount of computational resources

Q: Can I use analytical methods to find a particular solution to a differential equation?

A: Yes, you can use analytical methods such as separation of variables or integration by substitution to find a particular solution to a differential equation.

Q: What are the advantages of using analytical methods to find a particular solution to a differential equation?

A: The advantages of using analytical methods to find a particular solution to a differential equation include:

  • They are accurate
  • They are efficient
  • They can handle complex equations

Q: What are the disadvantages of using analytical methods to find a particular solution to a differential equation?

A: The disadvantages of using analytical methods to find a particular solution to a differential equation include:

  • They may be difficult to implement
  • They may require a large amount of mathematical knowledge
  • They may not be applicable to all types of differential equations

Q: Can I use a combination of numerical and analytical methods to find a particular solution to a differential equation?

A: Yes, you can use a combination of numerical and analytical methods to find a particular solution to a differential equation.

Q: What are the advantages of using a combination of numerical and analytical methods to find a particular solution to a differential equation?

A: The advantages of using a combination of numerical and analytical methods to find a particular solution to a differential equation include:

  • They can handle complex equations
  • They can provide accurate results
  • They can be efficient

Q: What are the disadvantages of using a combination of numerical and analytical methods to find a particular solution to a differential equation?

A: The disadvantages of using a combination of numerical and analytical methods to find a particular solution to a differential equation include:

  • They may be difficult to implement
  • They may require a large amount of mathematical knowledge
  • They may not be applicable to all types of differential equations