Show that each of the following differential equations is homogeneous and solve each of them :
`(x^(2)-y^(2))dx+2xy dy=0`

Answer

To solve the differential equation \((x^{2}-y^{2})dx + 2xy dy = 0\), we will follow these steps:

### Step 1: Rewrite the Equation
We start by rewriting the given differential equation in a more manageable form:
\[ (x^2 - y^2)dx + 2xy dy = 0 \]
This can be rearranged to:
\[ (x^2 - y^2)dx = -2xy dy \]

### Step 2: Separate Variables
We can express \(\frac{dy}{dx}\) in terms of \(x\) and \(y\):
\[ \frac{dy}{dx} = \frac{x^2 - y^2}{-2xy} \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{x^2 - y^2}{2xy} \]

### Step 3: Check for Homogeneity
To check if the equation is homogeneous, we substitute \(x = \lambda x\) and \(y = \lambda y\):
\[ \frac{dy}{dx} = -\frac{(\lambda x)^2 - (\lambda y)^2}{2(\lambda x)(\lambda y)} = -\frac{\lambda^2(x^2 - y^2)}{2\lambda^2 xy} = -\frac{x^2 - y^2}{2xy} \]
Since the form remains unchanged, the differential equation is homogeneous.

### Step 4: Substitute \(y = tx\)
Let \(y = tx\), where \(t\) is a function of \(x\). Then, we differentiate:
\[ \frac{dy}{dx} = t + x\frac{dt}{dx} \]
Substituting this into our equation gives:
\[ t + x\frac{dt}{dx} = -\frac{x^2 - (tx)^2}{2x(tx)} = -\frac{x^2(1 - t^2)}{2tx^2} = -\frac{1 - t^2}{2t} \]
This leads to:
\[ t + x\frac{dt}{dx} = -\frac{1 - t^2}{2t} \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ x\frac{dt}{dx} = -\frac{1 - t^2}{2t} - t \]
Combining the terms:
\[ x\frac{dt}{dx} = -\frac{1 - t^2 + 2t^2}{2t} = -\frac{1 + t^2}{2t} \]

### Step 6: Separate Variables Again
Now we separate variables:
\[ \frac{2t}{1 + t^2} dt = -\frac{dx}{x} \]

### Step 7: Integrate Both Sides
Integrating both sides:
\[ \int \frac{2t}{1 + t^2} dt = \int -\frac{dx}{x} \]
The left side integrates to \(\log(1 + t^2)\) and the right side integrates to \(-\log|x| + C\):
\[ \log(1 + t^2) = -\log|x| + C \]

### Step 8: Substitute Back for \(t\)
Recall that \(t = \frac{y}{x}\), so:
\[ \log(1 + \frac{y^2}{x^2}) = -\log|x| + C \]
Exponentiating both sides gives:
\[ 1 + \frac{y^2}{x^2} = \frac{C}{|x|} \]
Multiplying through by \(x^2\):
\[ y^2 + x^2 = Cx \]

### Final Solution
Thus, the solution to the differential equation is:
\[ x^2 + y^2 = Cx \]