Solve the following differential equations
`(x^3-3xy^2)dx=(y^3-3x^2y)dy`.

Answer

To solve the differential equation \((x^3 - 3xy^2)dx = (y^3 - 3x^2y)dy\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ (x^3 - 3xy^2)dx = (y^3 - 3x^2y)dy \]
We can rewrite this as:
\[ \frac{dy}{dx} = \frac{x^3 - 3xy^2}{y^3 - 3x^2y} \]

### Step 2: Check for homogeneity
The equation is homogeneous. A differential equation is homogeneous if it can be expressed in the form \(f(tx, ty) = t^n f(x, y)\) for some \(n\). In this case, both the numerator and denominator are homogeneous functions of degree 3.

### Step 3: Substitute \(y = vx\)
Let \(y = vx\), where \(v\) is a function of \(x\). Then, we have:
\[ dy = vdx + x \frac{dv}{dx} \]
Substituting \(y\) and \(dy\) into the equation gives:
\[ v + x \frac{dv}{dx} = \frac{x^3 - 3x(vx)^2}{(vx)^3 - 3x^2(vx)} \]
This simplifies to:
\[ v + x \frac{dv}{dx} = \frac{x^3(1 - 3v^2)}{x^3(v^3 - 3v)} \]
Cancelling \(x^3\) (assuming \(x \neq 0\)):
\[ v + x \frac{dv}{dx} = \frac{1 - 3v^2}{v^3 - 3v} \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ x \frac{dv}{dx} = \frac{1 - 3v^2}{v^3 - 3v} - v \]
This can be simplified to:
\[ x \frac{dv}{dx} = \frac{1 - 3v^2 - v(v^3 - 3v)}{v^3 - 3v} \]
Combining terms in the numerator:
\[ x \frac{dv}{dx} = \frac{1 - 3v^2 - v^4 + 3v^2}{v^3 - 3v} = \frac{1 - v^4}{v^3 - 3v} \]

### Step 5: Separate variables
Now, we can separate variables:
\[ \frac{v^3 - 3v}{1 - v^4} dv = \frac{dx}{x} \]

### Step 6: Integrate both sides
Integrating both sides:
\[ \int \frac{v^3 - 3v}{1 - v^4} dv = \int \frac{dx}{x} \]

### Step 7: Solve the integrals
The integral on the right side is straightforward:
\[ \int \frac{dx}{x} = \ln |x| + C \]
For the left side, we can use partial fraction decomposition or direct integration techniques, but we will focus on the result:
\[ \text{Let } u = 1 - v^4 \Rightarrow du = -4v^3 dv \]
This leads to:
\[ -\frac{1}{4} \int \frac{du}{u} = \ln |u| + C \]
Substituting back gives:
\[ -\frac{1}{4} \ln |1 - v^4| = \ln |x| + C \]

### Step 8: Substitute back for \(v\)
Recall that \(v = \frac{y}{x}\):
\[ -\frac{1}{4} \ln |1 - \left(\frac{y}{x}\right)^4| = \ln |x| + C \]
This can be rewritten as:
\[ \ln |1 - \frac{y^4}{x^4}| = -4 \ln |x| + C' \]

### Final Step: Solve for the general solution
Exponentiating both sides gives:
\[ |1 - \frac{y^4}{x^4}| = Kx^{-4} \]
where \(K = e^{C'}\). Rearranging gives us the general solution of the differential equation.