$\mathrm{If}y=3e^{2x}+2e^{3x},\mathrm{then}\frac{d^{2}y}{d{x}^2}+6y\mathrm{is}\mathrm{equal}\mathrm{to}$

The correct option is:
$$5\frac{dy}{dx}$$

Step-by-step Derivation:

We are given the function:
$$y=3e^{2x}+2e^{3x}$$
We need to find the value of the expression:
$$\frac{d^{2}y}{d{x}^2}+6y$$

First, let's find the first derivative of y with respect to x, denoted as $\frac{dy}{dx}$.
Using the chain rule for differentiation, where $\frac{d}{dx}\left(e^{ax}\right)=a{e}^{ax}$, we have:
$$\frac{dy}{dx}=\frac{d}{dx}(3e^{2x}+2e^{3x})$$
$$\frac{dy}{dx}=3\cdot \left(2e^{2x}\right)+2\cdot \left(3e^{3x}\right)$$
$$\frac{dy}{dx}=6e^{2x}+6e^{3x}$$

Next, let's find the second derivative of y with respect to x, denoted as $\frac{d^{2}y}{d{x}^2}$:
$$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}(6e^{2x}+6e^{3x})$$
$$\frac{d^{2}y}{d{x}^2}=6\cdot \left(2e^{2x}\right)+6\cdot \left(3e^{3x}\right)$$
$$\frac{d^{2}y}{d{x}^2}=12e^{2x}+18e^{3x}$$

Now, substitute the expressions for $\frac{d^{2}y}{d{x}^2}$ and y into the required expression:
$$\frac{d^{2}y}{d{x}^2}+6y=(12e^{2x}+18e^{3x})+6(3e^{2x}+2e^{3x})$$
Simplify by expanding the term with coefficient 6:
$$\frac{d^{2}y}{d{x}^2}+6y=12e^{2x}+18e^{3x}+18e^{2x}+12e^{3x}$$
Combine the like terms containing $e^{2x}$ and $e^{3x}$:
$$\frac{d^{2}y}{d{x}^2}+6y=(12+18)e^{2x}+(18+12)e^{3x}$$
$$\frac{d^{2}y}{d{x}^2}+6y=30e^{2x}+30e^{3x}$$

Factor out 5 from the right-hand side to relate it to the first derivative:
$$\frac{d^{2}y}{d{x}^2}+6y=5(6e^{2x}+6e^{3x})$$

Since we already established that $\frac{dy}{dx}=6e^{2x}+6e^{3x}$, we can substitute this back:
$$\frac{d^{2}y}{d{x}^2}+6y=5\frac{dy}{dx}$$