Let p in mathbb{R}. Then the differential equ | vedclass

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(B) Given the family of curves: $y=(\alpha+\beta x) e^{p x}$  $(i)$ Differentiating with respect to $x$: $y^{\prime} = \beta e^{p x} + p(\alpha+\beta

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$y^{\prime \prime}-2 p y^{\prime}+p^2 y=0$

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(B) Given the family of curves: $y=(\alpha+\beta x) e^{p x}$  $(i)$ Differentiating with respect to $x$: $y^{\prime} = \beta e^{p x} + p(\alpha+\beta x) e^{p x}$ $y^{\prime} = \beta e^{p x} + p y$  (ii) From (ii), we have $\beta e^{p x} = y^{\prime} - p y$. Differentiating (ii) again with respect to $x$: $y^{\prime \prime} = \beta p e^{p x} + p y^{\prime}$ Substitute $\beta e^{p x} = y^{\prime} - p y$ into the equation: $y^{\prime \prime} = p(y^{\prime} - p y) + p y^{\prime}$ $y^{\prime \prime} = p y^{\prime} - p^2 y + p y^{\prime}$ $y^{\prime \prime} = 2 p y^{\prime} - p^2 y$ Rearranging the terms, we get: $y^{\prime \prime} - 2 p y^{\prime} + p^2 y = 0$

Let $p \in \mathbb{R}$. Then the differential equation of the family of curves $y=(\alpha+\beta x) e^{p x}$, where $\alpha$ and $\beta$ are arbitrary constants, is

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