Solve the following differential equations
`x^2y dx -(x^3+y^3)dy=0`.

Answer

To solve the differential equation \( x^2y \, dx - (x^3 + y^3) \, dy = 0 \), we can follow these steps:

### Step 1: Rewrite the Differential Equation
We start with the given equation:
\[ x^2y \, dx - (x^3 + y^3) \, dy = 0 \]
We can rearrange it to express \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{x^2y}{x^3 + y^3} \]

### Step 2: Use Substitution
Let's use the substitution \( y = xt \), where \( t \) is a function of \( x \). Then, we have:
\[ dy = x \, dt + t \, dx \]
Substituting \( y \) and \( dy \) into the equation gives:
\[ \frac{dy}{dx} = \frac{x^2(xt)}{x^3 + (xt)^3} = \frac{x^3t}{x^3 + x^3t^3} = \frac{xt}{1 + t^3} \]
Now substituting \( dy \) into the equation:
\[ x \, dt + t \, dx = \frac{xt}{1 + t^3} \, dx \]

### Step 3: Rearranging the Equation
Rearranging gives:
\[ x \, dt = \left(\frac{xt}{1 + t^3} - t\right) dx \]
Factoring out \( t \):
\[ x \, dt = \left(\frac{xt - t(1 + t^3)}{1 + t^3}\right) dx \]
This simplifies to:
\[ x \, dt = \left(\frac{xt - t - t^4}{1 + t^3}\right) dx \]

### Step 4: Separate Variables
We can separate variables:
\[ \frac{1 + t^3}{t(t^3 + 1)} \, dt = -\frac{dx}{x} \]

### Step 5: Integrate Both Sides
Now we integrate both sides:
\[ \int \frac{1 + t^3}{t^4} \, dt = -\int \frac{dx}{x} \]
This gives:
\[ \int \left(t^{-4} + t^{-1}\right) \, dt = -\ln |x| + C \]
Calculating the integrals:
\[ -\frac{1}{3t^3} + \ln |t| = -\ln |x| + C \]

### Step 6: Substitute Back for \( t \)
Recall that \( t = \frac{y}{x} \):
\[ -\frac{1}{3\left(\frac{y}{x}\right)^3} + \ln\left(\frac{y}{x}\right) = -\ln |x| + C \]
This simplifies to:
\[ -\frac{x^3}{3y^3} + \ln y - \ln x = C - \ln |x| \]

### Step 7: Final Form
Rearranging gives us:
\[ \ln y = C + \frac{x^3}{3y^3} + \ln x \]
Thus, the final solution can be expressed as:
\[ \ln y = C + \frac{x^3}{3y^3} + \ln x \]