Question

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

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

Step-by-step solution

Identify the Given Equation

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

Apply the Variable Change

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

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: \[ y^{\bullet} = \frac{r r^{\bullet} - x}{\sqrt{r^2 - x^2}} \]

Transform the Given Equation

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

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: \[ y y^{\bullet 2} + 2x y^{\bullet} - y = \frac{r^2 r^{\bullet 2} - 2x r r^{\bullet} + x^2 + 2x r r^{\bullet} - 2x^2 - r^2 + x^2}{\sqrt{r^2 - x^2}} \]

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 \(r^2\): \[ r^2 (r^{\bullet 2} - 1) = 0 \]

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: \[ r^{\bullet} = \pm 1 \]

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.

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: \[ y = \sqrt{2xC + C^2} \]

Key concepts

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.