Help Engineering maths

Question

I am really stuck on this, I've done the first part need help on the second part. The potential, V, between two hollow, concentric metallic spheres is given by the solution of the following equation $$ \frac{1}{r^2}\frac{d}{dr}\left[r^2\left(\frac{dv}{dr}\right)\right]=0 $$ Show that the general solution of this equation is $$ V(r) =3 - \frac{2}{r} \,. $$ The spheres have radii of 1 m and 2 m. The voltage on the inner sphere is 1 V and the voltage on the outer sphere is 2V. The boundary conditions for the above differential equation are therefore $V(r=1)=1$ and $V(r=2)=2$. Use these conditions to show that the particular solution is $$ V(r) =3 - \frac{2}{r} \,. $$

Answer 1

Hint:

Note that you can greatly simplify the expression by computing $$ \frac{d}{dr}\left[r^2\frac{dV}{dr}\right]=r^2\frac{d}{dr}\left[\frac{dV}{dr}\right]+\frac{d}{dr}\left[r^2\right]\frac{dV}{dr}=r^2\frac{d^2V}{dr^2}+2r\frac{dV}{dr}, $$ making your differential equation $$ \frac{d^2V}{dr^2}+\frac{2}{r}\frac{dV}{dr}=0,\qquad\text{OR}\qquad V''(r)+\frac{2}{r}V'(r)=0 $$ Let $w(r)=V'(r)$. Then we see that $$ w'(r)+\frac{2}{r}w(r)=0, $$ or $$ w'(r)=-\frac{2}{r}w(r). $$ This is a separable equation, and can therefore be solved pretty directly! Then, you'll have an expression for $w=V'$ in terms of $r$, and can integrate to get back to $V$. All that's left is to use the conditions given in the problem to establish initial conditions, which will let you solve for the undetermined constants.