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Solve the differential equation \(y y^{\prime 2}+2 x y^{\prime}-y=0\) by changing from variables \(y, x\) to \(r, x,\) where \(y^{2}=r^{2}-x^{2} ;\) then \(y y^{\prime}=r r^{\prime}-x\).

Short Answer

Expert verified
The solution to the differential equation is \( y = \sqrt{2xC + C^2} \).

Step by step solution

01

Identify the Given Equation

The given differential equation is: \[ y y^{\bullet 2} + 2x y^{\bullet} - y = 0 \]
02

Apply the Variable Change

We change variables from \(y, x\) to \(r, x\) where \(y^2 = r^2 - x^2\). Therefore, \[ y = \sqrt{r^2 - x^2} \]
03

Differentiate the Transformed Variable

Using the derivative of \(y\), we get: \[ y^{\bullet} = \frac{d}{dx}(\sqrt{r^2 - x^2}) \] Applying the chain rule, this becomes: \[ y^{\bullet} = \frac{r r^{\bullet} - x}{\sqrt{r^2 - x^2}} \]
04

Transform the Given Equation

Replace \(y y^{\bullet}\) as per the given transformation: \[ y y^{\bullet} = r r^{\bullet} - x \] Thus, \[ y y^{\bullet 2} + 2x y^{\bullet} - y = (r r^{\bullet} - x)^2 + 2x \left( \frac{r r^{\bullet} - x}{\sqrt{r^2 - x^2}} \right) - \sqrt{r^2 - x^2} \]
05

Simplify the Transformed Equation

First, simplify \((r r^{\bullet} - x)^2\): \[ (r r^{\bullet} - x)^2 = r^2 r^{\bullet 2} - 2x r r^{\bullet} + x^2 \] Next, simplify the second term: \[ 2x \left( \frac{r r^{\bullet} - x}{\sqrt{r^2 - x^2}} \right) = \frac{2x r r^{\bullet} - 2x^2}{\sqrt{r^2 - x^2}} \] Combine and simplify: \[ y y^{\bullet 2} + 2x y^{\bullet} - y = \frac{r^2 r^{\bullet 2} - 2x r r^{\bullet} + x^2 + 2x r r^{\bullet} - 2x^2 - r^2 + x^2}{\sqrt{r^2 - x^2}} \]
06

Solve the Simplified Equation

Combine terms and solve the equation: \[ r^2 r^{\bullet 2} - 2x^2 - r^2 + x^2 = 0 \] Simplify further: \[ r^2 r^{\bullet 2} - r^2 = 0 \] Factor out \(r^2\): \[ r^2 (r^{\bullet 2} - 1) = 0 \]
07

Solve for the Derivative

Assuming \(r \eq 0\), the solution to the differential equation is: \[ r^{\bullet 2} = 1 \] Taking the square root of both sides: \[ r^{\bullet} = \pm 1 \]
08

Integrate to Find r

Integrate with respect to \(x\) to find \(r\): \[ \frac{dr}{dx} = \pm 1 \] Therefore, \[ r = \pm x + C \] where \(C\) is the constant of integration.
09

Substitute back to find y

Using the relationship \( y^2 = r^2 - x^2 \), substitute \( r = \pm x + C \): \[ y^2 = (\pm x + C)^2 - x^2 \] Simplify the expression: \[ y^2 = (x^2 + 2xC + C^2) - x^2 \] Thus, \[ y^2 = 2xC + C^2 \] As a result: \[ y = \sqrt{2xC + C^2} \]

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Key Concepts

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

Change of Variables
Change of variables is a method used to simplify differential equations by transforming them into a different set of variables. This often makes the equations easier to solve.
In this exercise, we change from variables \(y, x\) to \(r, x\), where we define \(y^2 = r^2 - x^2\).
We then express \(y\) and its derivative \(y'\) in terms of \(r\) and \(r'\).
Specifically, \(y \) can be written as \(\sqrt{r^2 - x^2}\) and \( y' \) can be transformed using the chain rule:
\( y' = \frac{r r' - x}{\sqrt{r^2 - x^2}} \)
This change of variables helps to simplify our differential equation into a more manageable form.
First-Order Differential Equations
Differential equations involving only the first derivative of the unknown function are called first-order differential equations.
These equations can often be solved using various methods including separation of variables, integrating factors, or, as in this case, change of variables.
The given equation \(y y'^{2} + 2x y' - y = 0 \) is a first-order differential equation.
By changing variables to \(r\) and \(x\), we transformed and simplified it to find the solution for \( r' \):
\[ r^2 (r'^2 - 1) = 0 \]
Here, we assumed \( r \eq 0 \) leading us to \( r'^{2} = 1 \).
This is easier to integrate and solve compared to the original form.
Integration
Integration is the process of finding the antiderivative or the original function from its derivative.
In solving our differential equation, after determining that \( r' = \pm 1 \), we integrate \( \frac{dr}{dx} = \pm 1 \) to find \( r \):
\[ r = \pm x + C \]
Here, \( C \) represents the constant of integration.
Once we have \( r \) in terms of \( x \), we can substitute back to find \( y \).
Using the relationship \( y^2 = r^2 - x^2 \) and substituting \( r = \pm x + C \), we get:
\[ y^2 = (\pm x + C)^2 - x^2 \]
Simplifying, it results in:
\[ y^2 = 2xC + C^2 \]
Therefore, we obtain the solution:
\[ y = \sqrt{2xC + C^2} \]
This entire process showcases the power of integration in solving differential equations.

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Most popular questions from this chapter

The force of gravitational attraction on a mass \(m\) at distance \(r\) from the center of the earth \((r>\text { radius } R \text { of the earth })\) is $$m g R^{2} / r^{2}$$. Then the differential equation of motion of a mass \(m\) projected radially outward from the surface of the earth, with initial velocity \(v_{0},\) is $$m d^{2} r / d t^{2}=-m g R^{2} / r^{2}$$. Use method (c) above to find \(v\) as a function of \(r\) if \(v=v_{0}\) initially (that is, when \(r=R) .\) Find the maximum value of \(r\) for a given \(v_{0},\) that is, the value of \(r\) when \(v=0 .\) Find the escape velocity, that is, the smallest value of \(v_{0}\) for which \(r\) can tend to infinity.

By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$y^{\prime \prime}+2 y^{\prime}+5 y=10 \cos t, \quad y_{0}=2, y_{0}^{\prime}=1$$

For each of the following differential equations, separate variables and find a solution containing one arbitrary constant. Then find the value of the constant to give a particular solution satisfying the given boundary condition. Computer plot a slope field and some of the solution curves. \(y^{\prime}+2 x y^{2}=0, \quad y=1\) when \(x=2\)

Solve the differential equation \(y^{\prime \prime}-a^{2} y=f(t),\) where \(f(t)=\left\\{\begin{array}{ll}0, & t<0 \\ 1, & t>0\end{array}\right.\) and \(y_{0}=y_{0}^{\prime}=0\). Hint: Use the convolution integral as in the example.

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$x^{2} y^{\prime}-x y=1 / x$$

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