Show that each of the following differential equations is homogeneous and solve each of them :
`(dy)/(dx)=(x^(2)y)/(x^(3)+y^(3))`

Answer

To solve the differential equation \(\frac{dy}{dx} = \frac{x^2 y}{x^3 + y^3}\), we will first show that it is homogeneous and then solve it.

### Step 1: Show that the equation is homogeneous
A differential equation is homogeneous if it can be expressed in the form \(\frac{dy}{dx} = f\left(\frac{y}{x}\right)\). We can check if our given equation is homogeneous by substituting \(x\) with \(\lambda x\) and \(y\) with \(\lambda y\):

\[ \frac{dy}{dx} = \frac{x^2 y}{x^3 + y^3} \]

Substituting \(x = \lambda x\) and \(y = \lambda y\):

\[ \frac{dy}{dx} = \frac{(\lambda x)^2 (\lambda y)}{(\lambda x)^3 + (\lambda y)^3} = \frac{\lambda^3 x^2 y}{\lambda^3 (x^3 + y^3)} = \frac{x^2 y}{x^3 + y^3} \]

Since the equation maintains its form after substitution, it confirms that the differential equation is homogeneous.

### Step 2: Substitute \(y = vx\)
Let \(y = vx\), where \(v\) is a function of \(x\). Then, we have:

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

Substituting \(y\) into the original equation gives:

\[ v + x\frac{dv}{dx} = \frac{x^2(vx)}{x^3 + (vx)^3} \]

This simplifies to:

\[ v + x\frac{dv}{dx} = \frac{vx^3}{x^3 + v^3 x^3} = \frac{v}{1 + v^3} \]

### Step 3: Rearranging the equation
Rearranging the equation, we have:

\[ x\frac{dv}{dx} = \frac{v}{1 + v^3} - v \]

This simplifies to:

\[ x\frac{dv}{dx} = v\left(\frac{1 - (1 + v^3)}{1 + v^3}\right) = -\frac{v^4}{1 + v^3} \]

### Step 4: Separate variables
Now, we separate the variables:

\[ \frac{1 + v^3}{v^4} dv = -\frac{1}{x} dx \]

### Step 5: Integrate both sides
Now, we integrate both sides:

\[ \int \frac{1 + v^3}{v^4} dv = -\int \frac{1}{x} dx \]

The left side becomes:

\[ \int \left(\frac{1}{v^4} + \frac{1}{v}\right) dv = -\ln|x| + C \]

Integrating gives:

\[ -\frac{1}{3v^3} + \ln|v| = -\ln|x| + C \]

### Step 6: Substitute back for \(y\)
Now, substituting back \(v = \frac{y}{x}\):

\[ -\frac{1}{3\left(\frac{y}{x}\right)^3} + \ln\left(\frac{y}{x}\right) = -\ln|x| + C \]

This can be rearranged and simplified to find the relationship between \(x\) and \(y\).

### Final Result
The final result will be in the form of an implicit equation relating \(x\) and \(y\).