Solve y dy/dx = x e^(x2 - 3y2)

The given equation is:

$$y \frac{dy}{dx} = x e^{x^2 - 3y^2}$$

We can rewrite the exponential term using the property $e^{a-b} = e^a e^{-b}$:

$$y \frac{dy}{dx} = x e^{x^2} e^{-3y^2}$$

To separate variables, multiply both sides by $e^{3y^2}$ and by $dx$:

$$y e^{3y^2} dy = x e^{x^2} dx$$

Let $u = 3y^2$. Then, differentiate $u$ with respect to $y$: $\frac{du}{dy} = 6y$.
So, $du = 6y \, dy$, which means $y \, dy = \frac{1}{6} du$.
Substitute $u$ and $y \, dy$ into the integral:

$$\int y e^{3y^2} dy = \int e^u \left(\frac{1}{6}\right) du = \frac{1}{6} \int e^u du = \frac{1}{6} e^u + C_1$$

Substitute back $u = 3y^2$:

$$\frac{1}{6} e^{3y^2} + C_1$$

Let $v = x^2$. Then, differentiate $v$ with respect to $x$: $\frac{dv}{dx} = 2x$.
So, $dv = 2x \, dx$, which means $x \, dx = \frac{1}{2} dv$.
Substitute $v$ and $x \, dx$ into the integral:

$$\int x e^{x^2} dx = \int e^v \left(\frac{1}{2}\right) dv = \frac{1}{2} \int e^v dv = \frac{1}{2} e^v + C_2$$

Substitute back $v = x^2$:

$$\frac{1}{2} e^{x^2} + C_2$$

$$\frac{1}{6} e^{3y^2} + C_1 = \frac{1}{2} e^{x^2} + C_2$$

Move the constants to one side, letting $C = C_2 - C_1$:

$$\frac{1}{6} e^{3y^2} = \frac{1}{2} e^{x^2} + C$$

Multiply the entire equation by 6 to clear the fractions:

$$e^{3y^2} = 3 e^{x^2} + 6C$$

Let $K = 6C$ be a new arbitrary constant:

$$e^{3y^2} = 3 e^{x^2} + K$$