Find a substitution that turns the following ODE into a linear one, then
solve it and give your answer to the initial ODE:
dy
dx − 2 y/x 
= x/y

We are given the nonlinear ODE:

\[
\frac{dy}{dx} - \frac{2y}{x} = \frac{x}{y}
\]

### Step 1: Analyze the structure of the equation

This is a first-order differential equation with a nonlinear term \(\frac{x}{y}\). To linearize it, we can try a substitution of the form \(v = y^2\), which is often helpful when dealing with terms like \(\frac{x}{y}\) or \(y\) in the denominator.

### Step 2: Perform the substitution

Let’s use the substitution \(v = y^2\), implying that:

\[
y = \sqrt{v}
\]

Now, differentiate \(v = y^2\) with respect to \(x\):

\[
\frac{dv}{dx} = 2y \frac{dy}{dx}
\]

Thus, we can express \(\frac{dy}{dx}\) as:

\[
\frac{dy}{dx} = \frac{1}{2y} \frac{dv}{dx}
\]

### Step 3: Substitute into the original ODE

Substitute \(\frac{dy}{dx}\) and the other terms involving \(y\) into the original equation. The original equation becomes:

\[
\frac{1}{2y} \frac{dv}{dx} - \frac{2y}{x} = \frac{x}{y}
\]

Now substitute \(y = \sqrt{v}\):

\[
\frac{1}{2\sqrt{v}} \frac{dv}{dx} - \frac{2\sqrt{v}}{x} = \frac{x}{\sqrt{v}}
\]

Multiply through by \(2\sqrt{v}\) to eliminate the denominators:

\[
\frac{dv}{dx} - \frac{4v}{x} = 2x
\]

This is now a linear differential equation in \(v\).

### Step 4: Solve the linear equation

We now have the linear equation:

\[
\frac{dv}{dx} - \frac{4v}{x} = 2x
\]

This is a first-order linear ODE, which we can solve using an integrating factor. The integrating factor \(\mu(x)\) is given by:

\[
\mu(x) = e^{\int -\frac{4}{x} dx} = e^{-4 \ln |x|} = \frac{1}{x^4}
\]

Multiplying through the equation by the integrating factor:

\[
\frac{1}{x^4} \frac{dv}{dx} - \frac{4v}{x^5} = \frac{2x}{x^4}
\]

This simplifies to:

\[
\frac{d}{dx} \left( \frac{v}{x^4} \right) = \frac{2}{x^3}
\]

### Step 5: Integrate both sides

Now integrate both sides with respect to \(x\):

\[
\frac{v}{x^4} = \int \frac{2}{x^3} dx = \frac{2}{-2x^2} = -\frac{1}{x^2}
\]

Thus, we have:

\[
v = -\frac{x^2}{x^4} + Cx^4 = -x^2 + Cx^4
\]

### Step 6: Substitute back \(v = y^2\)

Recall that \(v = y^2\), so:

\[
y^2 = -x^2 + Cx^4
\]

Taking the square root of both sides:

\[
y = \pm \sqrt{-x^2 + Cx^4}
\]

### Step 7: Final answer

The general solution to the original nonlinear ODE is:

\[
y(x) = \pm \sqrt{Cx^4 - x^2}
\]

where \(C\) is the constant of integration.

Would you like more details on any step, or clarification of any part of the process?

Here are some related questions for deeper exploration:

1. How do integrating factors help in solving linear ODEs?
2. What are common substitutions for nonlinear ODEs?
3. What is the role of the constant of integration in differential equations?
4. Can every nonlinear ODE be transformed into a linear one?
5. How does initial condition influence the solution to an ODE?

**Tip:** Substitutions often simplify complex ODEs by reducing nonlinearity or transforming the equation into a solvable standard form.

Find a substitution that turns the following ODE into a linear one, then solve it and give your answer to the initial ODE: dy/dx − 2y/x = x/y

Learn how to solve the nonlinear ordinary differential equation dy/dx - 2y/x = x/y by using substitution. This problem focuses on key calculus concepts like linearizing nonlinear ODEs and using integrating factors. The solution provides a step-by-step guide suitable for undergraduate-level calculus courses.

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