From the image, it appears there are two mathematical tasks outlined:

1. First equation: $y = C_1 e^{4x} + C_2 e^{-2x}$
   The task seems to involve finding the Differential Equation (D.E.) of the given general solution. This means deriving a differential equation for which this solution is valid.

2. Second equation: $3x^2 - xy^2 = C$
   It seems to be an implicit equation involving $x$, $y$, and $C$. The task might involve deriving some properties of this equation or analyzing it further.

I will analyze and solve each problem.

1. Derive the D.E. of $y = C_1 e^{4x} + C_2 e^{-2x}$:

1. Differentiate $y$ with respect to $x$: $y' = 4C_1 e^{4x} - 2C_2 e^{-2x}$
2. Differentiate $y'$ again to find $y''$: $y'' = 16C_1 e^{4x} + 4C_2 e^{-2x}$
3. Express $y''$, $y'$, and $y$ in terms of the original solution:
   • Notice that $e^{4x}$ and $e^{-2x}$ can be eliminated.
   • Combine terms into a single differential equation: $y'' + 2y' - 8y = 0$

So, the Differential Equation is: $y'' + 2y' - 8y = 0$

2. Analysis of $3x^2 - xy^2 = C$:

This represents a family of curves, depending on the constant $C$.

1. To analyze this equation, calculate the derivative $\frac{dy}{dx}$ implicitly:
   $\frac{d}{dx}(3x^2 - xy^2) = \frac{d}{dx}(C)$ Differentiating both sides: $6x - y^2 - 2xy\frac{dy}{dx} = 0$ Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{6x - y^2}{2xy}$

This gives the slope of the tangent line to the curve at any point $(x, y)$.

Would you like further steps or analysis for any of these equations?

Questions for Further Exploration:

1. What are the physical or geometric interpretations of the solution $y = C_1 e^{4x} + C_2 e^{-2x}$?
2. Can the solution $y = C_1 e^{4x} + C_2 e^{-2x}$ be solved for specific initial conditions?
3. What is the nature of the curve represented by $3x^2 - xy^2 = C$ (e.g., ellipse, hyperbola)?
4. How does $\frac{dy}{dx} = \frac{6x - y^2}{2xy}$ behave at certain points on the curve?
5. Are there critical points or symmetry properties of the equation $3x^2 - xy^2 = C$?

Tip:

When deriving differential equations from general solutions, always eliminate constants by taking derivatives and combining terms systematically.