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

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

Find the first derivative using $\frac{d}{dx}(e^{ax}) = a \cdot e^{ax}$:

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

$\frac{dy}{dx} = 3 \cdot 2e^{2x} + 2 \cdot 3e^{3x}$

$\frac{dy}{dx} = 6e^{2x} + 6e^{3x}$

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Find the second derivative:

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

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

$\frac{d^2y}{dx^2} = 12e^{2x} + 18e^{3x}$

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Calculate $\frac{d^2y}{dx^2} + 6y$:

$\frac{d^2y}{dx^2} + 6y = (12e^{2x} + 18e^{3x}) + 6(3e^{2x} + 2e^{3x})$

$= 12e^{2x} + 18e^{3x} + 18e^{2x} + 12e^{3x}$

$= (12 + 18)e^{2x} + (18 + 12)e^{3x}$

$= 30e^{2x} + 30e^{3x}$

$= 30(e^{2x} + e^{3x})$

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From the first derivative, factor out 6:

$\frac{dy}{dx} = 6e^{2x} + 6e^{3x} = 6(e^{2x} + e^{3x})$

Therefore:

$e^{2x} + e^{3x} = \frac{1}{6} \cdot \frac{dy}{dx}$

Substituting back:

$\frac{d^2y}{dx^2} + 6y = 30(e^{2x} + e^{3x})$

$= 30 \cdot \frac{1}{6} \cdot \frac{dy}{dx}$

$= \frac{30}{6} \cdot \frac{dy}{dx}$

$= 5\frac{dy}{dx}$

Therefore, $\frac{d^2y}{dx^2} + 6y = 5\frac{dy}{dx}$