Question

Show that \(V=(A / r)+B\) is a solution of the differential equation \(\frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dr}^{2}}+\frac{2}{\mathrm{r}} \frac{\mathrm{dV}}{\mathrm{dr}}=0\).

Short answer

Question: Prove that the function \(V = (A / r) + B\) is a solution to the given differential equation \(\frac{\mathrm{d^2V}}{\mathrm{dr^2}}+\frac{2}{\mathrm{r}}\frac{\mathrm{dV}}{\mathrm{dr}}=0\). Answer: We have found the first derivative of V with respect to r as \(\frac{\mathrm{dV}}{\mathrm{dr}} = - \frac{A}{r^2}\) and the second derivative as \(\frac{\mathrm{d^2V}}{\mathrm{dr^2}} = \frac{2A}{r^3}\). By substituting these derivatives back into the differential equation and simplifying the equation, we obtain \(0 = 0\), which confirms that the function \(V = (A / r) + B\) is a solution to the given differential equation.

Step-by-step solution

Find the first derivative of the function V with respect to r

To find the first derivative \(\frac{\mathrm{dV}}{\mathrm{dr}}\), we will use the power rule and the constant rule of derivatives. We have: \(\frac{\mathrm{dV}}{\mathrm{dr}} = \frac{\mathrm{d}}{\mathrm{dr}}(A / r) + \frac{\mathrm{d}}{\mathrm{dr}} B = - \frac{A}{r^2}\) The first derivative of V with respect to r is \(\frac{\mathrm{dV}}{\mathrm{dr}} = - \frac{A}{r^2}\).

Find the second derivative of the function V with respect to r

To find the second derivative \(\frac{\mathrm{d^2V}}{\mathrm{dr^2}}\), we will differentiate the first derivative with respect to r. Again, we will use the power rule: \(\frac{\mathrm{d^2V}}{\mathrm{dr^2}} = \frac{\mathrm{d}}{\mathrm{dr}}(- \frac{A}{r^2}) = \frac{2A}{r^3}\) The second derivative of V with respect to r is \(\frac{\mathrm{d^2V}}{\mathrm{dr^2}} = \frac{2A}{r^3}\).

Plug the derivatives into the differential equation

Now, we will substitute our first and second derivatives back into the given differential equation: \(\frac{\mathrm{d^2V}}{\mathrm{dr^2}}+\frac{2}{\mathrm{r}}\frac{\mathrm{dV}}{\mathrm{dr}}=0\) Plugging in our derivatives: \(\frac{2A}{r^3}+\frac{2}{r}(-\frac{A}{r^2})=0\)

Simplify the equation

Now, we will simplify the equation to see if it is true: \(\frac{2A}{r^3}+\frac{-2A}{r^3}=\frac{2A-2A}{r^3}=0\) The equation holds true, as the left side simplifies to 0, which equals the right side. Thus, the function \(V = (A / r) + B\) is a solution to the given differential equation \(\frac{\mathrm{d^2V}}{\mathrm{dr^2}}+\frac{2}{\mathrm{r}}\frac{\mathrm{dV}}{\mathrm{dr}}=0\).