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

Answer

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

### Step 1: Rewrite the Differential Equation
We start by rewriting the given equation in the standard form:
\[ \frac{dy}{dx} = \frac{x^2 + y^2}{2xy} \]

### Step 2: Check for Homogeneity
A function \(f(x, y)\) is homogeneous of degree \(n\) if \(f(\lambda x, \lambda y) = \lambda^n f(x, y)\) for any non-zero scalar \(\lambda\).

In our case:
\[ f(x, y) = \frac{x^2 + y^2}{2xy} \]
Substituting \(\lambda x\) and \(\lambda y\):
\[ f(\lambda x, \lambda y) = \frac{(\lambda x)^2 + (\lambda y)^2}{2(\lambda x)(\lambda y)} = \frac{\lambda^2(x^2 + y^2)}{2\lambda^2 xy} = \frac{x^2 + y^2}{2xy} = f(x, y) \]
This shows that \(f(x, y)\) is homogeneous of degree 0.

### Step 3: Use the Substitution \(y = vx\)
We will use the substitution \(y = vx\), where \(v\) is a function of \(x\). Then, we have:
\[ \frac{dy}{dx} = v + x\frac{dv}{dx} \]
Substituting \(y\) in the equation gives:
\[ \frac{dy}{dx} = v + x\frac{dv}{dx} = \frac{x^2 + (vx)^2}{2x(vx)} = \frac{x^2(1 + v^2)}{2vx^2} = \frac{1 + v^2}{2v} \]

### Step 4: Rearranging the Equation
Now, we can rearrange the equation:
\[ v + x\frac{dv}{dx} = \frac{1 + v^2}{2v} \]
This simplifies to:
\[ x\frac{dv}{dx} = \frac{1 + v^2}{2v} - v \]
\[ x\frac{dv}{dx} = \frac{1 + v^2 - 2v^2}{2v} = \frac{1 - v^2}{2v} \]

### Step 5: Separate Variables
Now we separate the variables:
\[ \frac{2v}{1 - v^2} dv = \frac{1}{x} dx \]

### Step 6: Integrate Both Sides
Integrating both sides:
\[ \int \frac{2v}{1 - v^2} dv = \int \frac{1}{x} dx \]
The left side can be integrated using the substitution \(u = 1 - v^2\), giving:
\[ -\ln|1 - v^2| = \ln|x| + C \]

### Step 7: Solve for \(v\)
Exponentiating both sides:
\[ |1 - v^2| = \frac{K}{x} \quad \text{(where \(K = e^{-C}\))} \]
Thus,
\[ 1 - v^2 = \frac{K}{x} \]
Substituting back \(v = \frac{y}{x}\):
\[ 1 - \left(\frac{y}{x}\right)^2 = \frac{K}{x} \]
This leads to:
\[ 1 - \frac{y^2}{x^2} = \frac{K}{x} \]
Multiplying through by \(x^2\) gives:
\[ x^2 - y^2 = Kx \]

### Final Solution
The general solution of the differential equation is:
\[ x^2 - y^2 - Kx = 0 \]