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

Answer

To solve the differential equation \((x+y)dy - (y-x)dx = 0\), we will first show that it is homogeneous and then solve it step by step.

### Step 1: Show that the differential equation is homogeneous

A differential equation is homogeneous if it can be expressed in the form \(f(\lambda x, \lambda y) = \lambda^n f(x, y)\) for some integer \(n\).

We can rewrite the given equation as:
\[ (x+y)dy = (y-x)dx \]
Dividing both sides by \(dx\), we have:
\[ \frac{dy}{dx} = \frac{y-x}{x+y} \]

Now, we will check if this function is homogeneous. We can substitute \(\lambda x\) and \(\lambda y\) into the right-hand side:
\[ f(\lambda x, \lambda y) = \frac{\lambda y - \lambda x}{\lambda x + \lambda y} = \frac{\lambda(y - x)}{\lambda(x + y)} = \frac{y - x}{x + y} \]
This shows that:
\[ f(\lambda x, \lambda y) = \lambda^0 f(x, y) \]
Thus, the function is homogeneous of degree 0.

### Step 2: Solve the differential equation

Since the equation is homogeneous, we can use the substitution \(y = zx\), where \(z = \frac{y}{x}\). Then, we have:
\[ dy = zdx + xdz \]
Substituting \(y\) and \(dy\) into the equation gives:
\[ (x + zx)(zdx + xdz) = (zx - x)dx \]
Simplifying this:
\[ (x(1 + z))(zdx + xdz) = (z - 1)xdx \]
Expanding both sides:
\[ xz(1 + z)dx + x^2(1 + z)dz = (z - 1)xdx \]
Dividing through by \(x\) (assuming \(x \neq 0\)):
\[ z(1 + z)dx + x(1 + z)dz = (z - 1)dx \]
Rearranging gives:
\[ x(1 + z)dz = (z - 1 - z(1 + z))dx \]
This simplifies to:
\[ x(1 + z)dz = (-1 - z^2)dx \]
Now we can separate variables:
\[ \frac{(1 + z)}{(-1 - z^2)}dz = \frac{1}{x}dx \]

### Step 3: Integrate both sides

Integrating both sides:
\[ \int \frac{(1 + z)}{(-1 - z^2)}dz = \int \frac{1}{x}dx \]
The right-hand side integrates to:
\[ \ln |x| + C \]
For the left-hand side, we can split the integral:
\[ \int \frac{1}{(-1 - z^2)}dz + \int \frac{z}{(-1 - z^2)}dz \]
The first integral can be solved using the arctangent function:
\[ -\frac{1}{\sqrt{1}} \tan^{-1}(z) = -\tan^{-1}(z) \]
The second integral can be solved using substitution:
\[ -\frac{1}{2} \ln |1 + z^2| \]
Thus, we have:
\[ -\tan^{-1}(z) - \frac{1}{2} \ln |1 + z^2| = \ln |x| + C \]

### Step 4: Substitute back for \(z\)

Recall that \(z = \frac{y}{x}\). Substituting back gives:
\[ -\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \ln \left|1 + \left(\frac{y}{x}\right)^2\right| = \ln |x| + C \]

### Final Solution

Thus, the solution to the differential equation \((x+y)dy - (y-x)dx = 0\) is:
\[ -\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \ln \left|1 + \frac{y^2}{x^2}\right| = \ln |x| + C \]