If the family of curves y = a e^{4x} + b e^{- | vedclass

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(B) Given the family of curves: $y = ae^{4x} + be^{-x}$ ... $(i)$Differentiating with respect to $x$:$\frac{dy}{dx} = 4ae^{4x} - be^{-x}$ ... (ii)Diff

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$\frac{d^3y}{dx^3} - 3\frac{d^2y}{dx^2} - 4\frac{dy}{dx}$

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(B) Given the family of curves: $y = ae^{4x} + be^{-x}$ ... $(i)$Differentiating with respect to $x$:$\frac{dy}{dx} = 4ae^{4x} - be^{-x}$ ... (ii)Differentiating again with respect to $x$:$\frac{d^2y}{dx^2} = 16ae^{4x} + be^{-x}$ ... (iii)To eliminate the constants $a$ and $b$, we can solve the system of linear equations $(i)$, (ii), and (iii). Adding $(i)$ and (ii):$y + \frac{dy}{dx} = 5ae^{4x} \implies ae^{4x} = \frac{1}{5}(y + \frac{dy}{dx})$Subtracting (ii) from $(i)$ multiplied by $4$:$4y - \frac{dy}{dx} = 5be^{-x} \implies be^{-x} = \frac{1}{5}(4y - \frac{dy}{dx})$Substituting these into (iii):$\frac{d^2y}{dx^2} = 16[\frac{1}{5}(y + \frac{dy}{dx})] + \frac{1}{5}(4y - \frac{dy}{dx})$$\frac{d^2y}{dx^2} = \frac{16y}{5} + \frac{16}{5}\frac{dy}{dx} + \frac{4y}{5} - \frac{1}{5}\frac{dy}{dx} = 4y + 3\frac{dy}{dx}$Thus, the differential equation is $f = \frac{d^2y}{dx^2} - 3\frac{dy}{dx} - 4y = 0$.To find $\frac{df}{dx}$, differentiate $f$ with respect to $x$:$\frac{df}{dx} = \frac{d}{dx}(\frac{d^2y}{dx^2} - 3\frac{dy}{dx} - 4y) = \frac{d^3y}{dx^3} - 3\frac{d^2y}{dx^2} - 4\frac{dy}{dx}$.

If the family of curves $y = a e^{4x} + b e^{-x}$, where $a, b$ are arbitrary constants, represents the general solution of the differential equation $f(x, y, \frac{dy}{dx}, \frac{d^2y}{dx^2}) = 0$, then find $\frac{df}{dx}$.

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