To solve the differential equation
\[
\frac{dx}{dy} = \frac{x + 2yx^2}{y - 2x^3}
\]
and find the curve that passes through the point (1, 0), we can follow these steps:
### Step 1: Rearranging the Equation
We start by rearranging the given differential equation. We can cross-multiply to obtain:
\[
dx \cdot (y - 2x^3) = (x + 2yx^2) \cdot dy
\]
### Step 2: Rearranging Terms
Next, we rearrange the equation to isolate terms involving \(dx\) and \(dy\):
\[
y \, dx - x \, dy = 2x^3 \, dx + 2yx^2 \, dy
\]
### Step 3: Collecting Like Terms
Now, we can collect the terms involving \(dx\) and \(dy\):
\[
y \, dx - 2x^3 \, dx = x \, dy + 2yx^2 \, dy
\]
This simplifies to:
\[
(y - 2x^3) \, dx = (x + 2yx^2) \, dy
\]
### Step 4: Dividing by \(x^2\)
Next, we divide both sides by \(x^2\):
\[
\frac{y - 2x^3}{x^2} \, dx = \left(\frac{x}{x^2} + 2y\right) \, dy
\]
This gives us:
\[
\frac{y}{x^2} - 2x \, dx = \left(\frac{1}{x} + 2y\right) \, dy
\]
### Step 5: Substituting \(z = \frac{y}{x}\)
Let \(z = \frac{y}{x}\). Then, we can express \(y\) as \(y = zx\) and differentiate:
\[
\frac{dz}{dx} = \frac{x \frac{dy}{dx} - y}{x^2}
\]
Substituting \(y = zx\) into the equation gives us:
\[
\frac{dz}{dx} = \frac{x \frac{dy}{dx} - zx}{x^2}
\]
### Step 6: Replacing in the Equation
We can replace \(y\) in the original equation with \(zx\) and simplify:
\[
\frac{dz}{dx} = -\frac{y \, dx - x \, dy}{x^2}
\]
### Step 7: Integrating
Now we can integrate both sides. The left side will integrate to:
\[
\int (2x \, dx + 2y \, dy) = x^2 + y^2 + \frac{y}{x} + C
\]
### Step 8: Applying Initial Condition
Now we apply the initial condition \( (1, 0) \):
\[
1^2 + 0^2 + \frac{0}{1} = C \implies C = 1
\]
### Final Step: Writing the Equation
Thus, the equation of the curve is:
\[
x^2 + y^2 + \frac{y}{x} = 1
\]
### Conclusion
The curve satisfying the differential equation and passing through the point (1, 0) is
\[
x^2 + y^2 + \frac{y}{x} = 1
\]
