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 equation in a more manageable form:
\[ (x^2 + y^2)dx + 2xy dy = 0 \]
This can be rearranged to:
\[ (x^2 + y^2)dx = -2xy dy \]
Dividing both sides by \(dx\) gives:
\[ 1 = -\frac{2xy}{x^2 + y^2} \frac{dy}{dx} \]
Thus, we can express \(\frac{dy}{dx}\) as:
\[ \frac{dy}{dx} = -\frac{x^2 + y^2}{2xy} \]

### Step 2: Check for homogeneity
To check if the equation is homogeneous, we can substitute \(y = vx\) (where \(v\) is a function of \(x\)). Then, we have:
\[ dy = v dx + x dv \]
Substituting \(y\) and \(dy\) into the equation gives:
\[ \frac{dy}{dx} = v + x\frac{dv}{dx} \]
Now, substituting \(y = vx\) into the right-hand side:
\[ \frac{dy}{dx} = -\frac{x^2 + (vx)^2}{2x(vx)} = -\frac{x^2(1 + v^2)}{2vx^2} = -\frac{1 + v^2}{2v} \]
Setting the two expressions for \(\frac{dy}{dx}\) equal:
\[ v + x\frac{dv}{dx} = -\frac{1 + v^2}{2v} \]
This confirms that the differential equation is homogeneous.

### Step 3: Substitute and simplify
Now we substitute \(y = vx\) into the equation:
\[ v + x\frac{dv}{dx} = -\frac{1 + v^2}{2v} \]
Rearranging gives:
\[ x\frac{dv}{dx} = -\frac{1 + v^2}{2v} - v \]
Combining the terms on the right:
\[ x\frac{dv}{dx} = -\frac{1 + v^2 + 2v^2}{2v} = -\frac{1 + 3v^2}{2v} \]

### Step 4: Separate variables
We can separate variables:
\[ \frac{2v}{1 + 3v^2} dv = -\frac{1}{x} dx \]

### Step 5: Integrate both sides
Integrating both sides:
\[ \int \frac{2v}{1 + 3v^2} dv = -\int \frac{1}{x} dx \]
For the left side, we can use substitution \(u = 1 + 3v^2\), \(du = 6v dv\):
\[ \frac{1}{3} \ln |1 + 3v^2| = -\ln |x| + C \]

### Step 6: Solve for \(v\)
Exponentiating both sides:
\[ |1 + 3v^2|^{1/3} = \frac{C}{|x|} \]
Substituting back \(v = \frac{y}{x}\):
\[ |1 + 3\left(\frac{y}{x}\right)^2|^{1/3} = \frac{C}{|x|} \]
This leads to:
\[ 1 + 3\frac{y^2}{x^2} = \frac{C^3}{x^3} \]
Multiplying through by \(x^3\):
\[ x^3 + 3y^2x = C^3 \]

### Final Solution
Thus, the general solution of the differential equation is:
\[ x^3 + 3xy^2 = C \]