If y = 3e2x + 2e3x, then $\frac{d^2y}{dx^2} + 6y$ is equal to
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For problems of this type, which relate a function to its derivatives, recognize the structure of a linear homogeneous differential equation. The function \(y = C_1e^{r_1x} + C_2e^{r_2x}\) is the solution to \(y'' - (r_1+r_2)y' + r_1r_2y = 0\). Here, \(r_1=2\) and \(r_2=3\), so it's a solution to \(y'' - 5y' + 6y = 0\). From this, we can directly see that \(y'' + 6y = 5y'\).
Step 1: Problem Overview: This problem requires calculating the first and second derivatives of a given function and then evaluating a specific expression involving these derivatives and the original function. Step 2: Relevant Calculus Rule: The primary differentiation rule to be applied is for exponential functions: \[ \frac{d}{dx}(e^{ax}) = a e^{ax} \]Step 3: Derivation and Evaluation: Given the function \(y = 3e^{2x} + 2e^{3x}\). Calculate the first derivative, \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{d}{dx}(3e^{2x} + 2e^{3x}) = 3(2e^{2x}) + 2(3e^{3x}) \]\[ \frac{dy}{dx} = 6e^{2x} + 6e^{3x} \]Calculate the second derivative, \(\frac{d^2y}{dx^2}\): \[ \frac{d^2y}{dx^2} = \frac{d}{dx}(6e^{2x} + 6e^{3x}) = 6(2e^{2x}) + 6(3e^{3x}) \]\[ \frac{d^2y}{dx^2} = 12e^{2x} + 18e^{3x} \]Now, substitute these into the expression \(\frac{d^2y}{dx^2} + 6y\): \[ \frac{d^2y}{dx^2} + 6y = (12e^{2x} + 18e^{3x}) + 6(3e^{2x} + 2e^{3x}) \]\[ = 12e^{2x} + 18e^{3x} + 18e^{2x} + 12e^{3x} \]Combine like terms: \[ = (12 + 18)e^{2x} + (18 + 12)e^{3x} \]\[ = 30e^{2x} + 30e^{3x} \]To relate this to the first derivative, factor out 5: \[ 30e^{2x} + 30e^{3x} = 5(6e^{2x} + 6e^{3x}) \]Recognize that the expression in the parentheses is \(\frac{dy}{dx}\): \[ \frac{d^2y}{dx^2} + 6y = 5\left(\frac{dy}{dx}\right) \]Step 4: Conclusion: The expression \(\frac{d^2y}{dx^2} + 6y\) simplifies to $5\frac{dy{dx}$}.