If y=3 e^{5 x}+5 e^{3 x},then frac{d^{2} y}{d | vedclass

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(C) Given $y = 3 e^{5 x} + 5 e^{3 x}$.First,find the first derivative $\frac{d y}{d x}$:$\frac{d y}{d x} = \frac{d}{d x}(3 e^{5 x} + 5 e^{3 x}) = 3(5

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(C) Given $y = 3 e^{5 x} + 5 e^{3 x}$.First,find the first derivative $\frac{d y}{d x}$:$\frac{d y}{d x} = \frac{d}{d x}(3 e^{5 x} + 5 e^{3 x}) = 3(5 e^{5 x}) + 5(3 e^{3 x}) = 15 e^{5 x} + 15 e^{3 x}$.Next,find the second derivative $\frac{d^{2} y}{d x^{2}}$:$\frac{d^{2} y}{d x^{2}} = \frac{d}{d x}(15 e^{5 x} + 15 e^{3 x}) = 15(5 e^{5 x}) + 15(3 e^{3 x}) = 75 e^{5 x} + 45 e^{3 x}$.Now,calculate $\frac{d^{2} y}{d x^{2}} - 8 \frac{d y}{d x}$:$\frac{d^{2} y}{d x^{2}} - 8 \frac{d y}{d x} = (75 e^{5 x} + 45 e^{3 x}) - 8(15 e^{5 x} + 15 e^{3 x})$$= 75 e^{5 x} + 45 e^{3 x} - 120 e^{5 x} - 120 e^{3 x}$$= -45 e^{5 x} - 75 e^{3 x}$$= -15(3 e^{5 x} + 5 e^{3 x})$$= -15 y$.

If $y=3 e^{5 x}+5 e^{3 x}$,then $\frac{d^{2} y}{d x^{2}}-8 \frac{d y}{d x}=$ (in $y$)

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