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| Eliminating the arbitrary constants $A$ and $B$ from the equation $v=\frac{A}{r}+B$ by differentiating what differential equation we will get? |
$\frac{d^2v}{dr^2}+\frac{dv}{dr}=0$ $\frac{d^2v}{dr^2}+\frac{1}{r}\frac{dv}{dr}=0$ $\frac{d^2v}{dr^2}+\frac{2}{r}\frac{dv}{dr}=0$ $\frac{d^2v}{dr^2}+2\frac{dv}{dr}=0$ |
| $\frac{d^2v}{dr^2}+\frac{2}{r}\frac{dv}{dr}=0$ |
| $v=\frac{A}{r}+B$. Differentiating w.r.to r we get $\frac{dv}{dr}=-\frac{A}{r^2}$. Differentiating again we get $\frac{d^2v}{dr^2}=\frac{2A}{r^3}$. Hence $\frac{r^3}{2}\frac{d^2v}{dr^2}=A=-r^2\frac{dv}{dr}$. So $\frac{r^3}{2}\frac{d^2v}{dr^2}+r^2\frac{dv}{dr}=0$. So we get $\frac{d^2v}{dr^2}+\frac{2}{r}\frac{dv}{dr}=0$ |
