The differential equation of the family of cu | vedclass

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(C) Given the equation $v = \frac{A}{r} + B$. Differentiating with respect to $r$,we get: $\frac{dv}{dr} = -\frac{A}{r^2}$. Differentiating again with

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$\frac{d^2v}{dr^2} + \frac{2}{r}\frac{dv}{dr} = 0$

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(C) Given the equation $v = \frac{A}{r} + B$. Differentiating with respect to $r$,we get: $\frac{dv}{dr} = -\frac{A}{r^2}$. Differentiating again with respect to $r$,we get: $\frac{d^2v}{dr^2} = \frac{2A}{r^3}$. From the first derivative,we have $A = -r^2 \frac{dv}{dr}$. Substituting this into the second derivative: $\frac{d^2v}{dr^2} = \frac{2(-r^2 \frac{dv}{dr})}{r^3} = -\frac{2}{r} \frac{dv}{dr}$. Rearranging the terms,we get: $\frac{d^2v}{dr^2} + \frac{2}{r} \frac{dv}{dr} = 0$.

The differential equation of the family of curves $v = \frac{A}{r} + B$,where $A$ and $B$ are arbitrary constants,is

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