We are given the differential equation:

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

The goal is to find an expression for the second derivative, $\frac{d^2y}{dx^2}$, in terms of $x$ and $y$.

Step-by-Step Solution:

1. Differentiate both sides of the equation $\frac{dy}{dx} = 3x^2 y$ with respect to $x$:

   $\frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}\left(3x^2 y\right).$

   The left-hand side becomes the second derivative:

   $\frac{d^2y}{dx^2}.$

2. Apply the product rule on the right-hand side:

   The product rule states that $\frac{d}{dx}(uv) = u'v + uv'$, where $u = 3x^2$ and $v = y$.

   Therefore:

   $\frac{d}{dx}(3x^2 y) = \frac{d}{dx}(3x^2) \cdot y + 3x^2 \cdot \frac{dy}{dx}.$

3. Differentiate $3x^2$ and substitute known values:

   $\frac{d}{dx}(3x^2) = 6x.$

   So:

   $\frac{d^2y}{dx^2} = 6x \cdot y + 3x^2 \cdot \frac{dy}{dx}.$

   But we already know that $\frac{dy}{dx} = 3x^2 y$. Substitute this into the equation:

   $\frac{d^2y}{dx^2} = 6x \cdot y + 3x^2 \cdot (3x^2 y).$

4. Simplify the expression:

   $\frac{d^2y}{dx^2} = 6x \cdot y + 9x^4 \cdot y.$

   Factor out $y$:

   $\frac{d^2y}{dx^2} = y(6x + 9x^4).$

Thus, the expression for the second derivative in terms of $x$ and $y$ is:

$\frac{d^2y}{dx^2} = y(6x + 9x^4).$

Would you like further details or have any questions?

Related Questions:

1. What is the general solution to the differential equation $\frac{dy}{dx} = 3x^2y$?
2. How would the solution change if $\frac{dy}{dx} = 3x^2 + y$?
3. Can this second derivative equation be solved for specific initial conditions?
4. What is the physical interpretation of this differential equation if $y$ represents position?
5. How does the second derivative behave as $x$ increases?

Tip:

Always remember to check for opportunities to use substitution from the original equation when differentiating implicit equations. This simplifies the algebra significantly.