(2xy)dx+(y^2-x^2)dy = 0 solve using homogenous equation

To solve the differential equation \((2xy)dx + (y^2 - x^2)dy = 0\) using the method of homogeneous equations, follow these steps:

### Step 1: Express the equation in standard form
First, rewrite the equation in a more convenient form:

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

### Step 2: Make the equation homogeneous
We check if the equation is homogeneous by substituting \(y = vx\) where \(v = \frac{y}{x}\) and then differentiating \(y = vx\) with respect to \(x\):

\[
\frac{dy}{dx} = v + x\frac{dv}{dx}
\]

Substitute \(y = vx\) and \(\frac{dy}{dx} = v + x\frac{dv}{dx}\) into the equation:

\[
v + x\frac{dv}{dx} = -\frac{2x(vx)}{(v^2x^2 - x^2)}
\]

Simplify this expression:

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

### Step 3: Separate variables
Rearrange the equation to separate variables:

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

\[
\frac{dv}{dx} = \frac{-v(v^2 - 1) - 2v}{x(v^2 - 1)}
\]

Simplify further:

\[
\frac{dv}{dx} = \frac{-v(v^2 + 1)}{x(v^2 - 1)}
\]

\[
\frac{v^2 - 1}{v(v^2 + 1)} dv = -\frac{dx}{x}
\]

### Step 4: Integrate both sides
Now, integrate both sides:

\[
\int \frac{v^2 - 1}{v(v^2 + 1)} dv = -\int \frac{dx}{x}
\]

The integral on the left side can be solved by splitting the fraction:

\[
\int \left(\frac{v^2}{v(v^2 + 1)} - \frac{1}{v(v^2 + 1)}\right) dv
\]

This simplifies to:

\[
\int \left(\frac{v}{v^2 + 1} - \frac{1}{v(v^2 + 1)}\right) dv
\]

These integrals can be solved individually:

\[
\int \frac{v}{v^2 + 1} dv - \int \frac{1}{v(v^2 + 1)} dv
\]

### Step 5: Solve the integrals
For the first integral:

\[
\int \frac{v}{v^2 + 1} dv = \frac{1}{2} \ln(v^2 + 1)
\]

For the second integral, use substitution \( u = v^2 + 1 \):

\[
\int \frac{1}{v(v^2 + 1)} dv
\]

The full solution becomes:

\[
\frac{1}{2} \ln(v^2 + 1) - \ln|v| = -\ln|x| + C
\]

### Step 6: Back-substitute \( v = \frac{y}{x} \)
Substitute back \( v = \frac{y}{x} \):

\[
\frac{1}{2} \ln\left(\frac{y^2}{x^2} + 1\right) - \ln\left|\frac{y}{x}\right| = -\ln|x| + C
\]

Simplify the expression:

\[
\ln\left(\frac{y^2}{x^2} + 1\right)^{\frac{1}{2}} - \ln\left|\frac{y}{x}\right| = -\ln|x| + C
\]

Finally, this can be written as:

\[
\ln\left(\frac{\sqrt{y^2 + x^2}}{y}\right) = \ln\left(\frac{1}{x}\right) + C
\]

Exponentiating both sides gives the general solution:

\[
\frac{\sqrt{y^2 + x^2}}{y} = \frac{K}{x}
\]

Where \(K\) is the constant.

### Final General Solution:
\[
\sqrt{y^2 + x^2} = Ky
\]

Or, equivalently:

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

### Do you want further details or have any questions?

Here are 5 related questions to consider:
1. How do you identify if a differential equation is homogeneous?
2. What is the substitution method used for solving homogeneous equations?
3. Can all homogeneous equations be solved using this method?
4. How do you verify the solution to a differential equation?
5. What other methods can be used to solve first-order differential equations?

**Tip:** When dealing with homogeneous equations, always consider substitutions that simplify the equation to separable form.

Learn how to solve the homogeneous differential equation (2xy)dx + (y^2 - x^2)dy = 0 using the method of homogeneous equations and separation of variables. This problem involves differential equations, homogeneous forms, and separation techniques. Suitable for college-level mathematics.

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