If y = (1 + alpha + alpha^2 + ldots) e^{nx},w | vedclass

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(C) Given,$y = (1 + \alpha + \alpha^2 + \ldots) e^{nx}$.Let $K = (1 + \alpha + \alpha^2 + \ldots)$,which is a constant.So,$y = K e^{nx}$.Differentiati

Vedclass · Accepted Answer

$n 	imes (	ext{error in } x)$

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(C) Given,$y = (1 + \alpha + \alpha^2 + \ldots) e^{nx}$.Let $K = (1 + \alpha + \alpha^2 + \ldots)$,which is a constant.So,$y = K e^{nx}$.Differentiating both sides with respect to $x$,we get:$\frac{dy}{dx} = K \cdot n e^{nx} = n \cdot (K e^{nx}) = ny$.Using the concept of differentials,$\Delta y \approx \frac{dy}{dx} \Delta x$.Therefore,$\Delta y = ny \Delta x$.Dividing both sides by $y$,we get the relative error in $y$:$\frac{\Delta y}{y} = n \Delta x$.Thus,the relative error in $y$ is $n 	imes (	ext{error in } x)$.

If $y = (1 + \alpha + \alpha^2 + \ldots) e^{nx}$,where $\alpha$ and $n$ are constants,then the relative error in $y$ is

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