$$\dfrac {d^2u}{dr^2}+\dfrac 1r\dfrac {du}{dr}-\dfrac u{r^2}+kr=0$$

Multiply by $r^2$ through out

$$\therefore r^2\dfrac {d^2u}{dr}+\dfrac {rdu}{dr}-U+kr^3=0------ (1)$$ This is a Cauchy homogenous different equations $$\therefore Let \space z=\space Pogo \space r $$ $ \therefore r=e^z \\ \therefore r^2\dfrac {d^2u}{dr}=D(D-1)U \\ r\dfrac {du}{dr}=Du \\ \therefore equation \space 1 \space becomes \\ D(D-1)U+DU-U=-ke^{3z}\\ \therefore (D^2-D+D-1)U=ke^{3z}\\ \therefore (D^2-1)U=-ke^{3z}\\ \text{Auxiliary equation is} D^2-1=0\\ D=\pm 1\\ Y_c=C_1e^{z}+C_2e^{-z}\\ P.I.\space is y_p=\dfrac 1{f(D)}x\\Y_p =\dfrac 1{D^2-1}-ke^{3z}\\ =-ke^{3z}\Bigg(\dfrac 1{3^2-1}\Bigg)\\ =\dfrac {-ke^{3z}}8 \\ \text{Solution is} y=Y_c+Y_p\\ =C_1e^z+C_2e^{-z}-\dfrac {ke^{3z}}8$

Resubstituting value of z we get

$y=C_1r-+\dfrac {C_2}r-\dfrac {kr^3}8$

It is also given that when

$$r=0,u=0, i.e., y=0\ \therefore D=0+\dfrac {C_2}r-0$$

This equation is valid only when $C_2=0\\ \therefore U=C_1r-\dfrac k8 r^2----- (1)$

Also given when $r=a,U=0\\ =C_1\times a-\dfrac k8a^3\\ C_1=\dfrac k8 a^2\\ \therefore y=U=\dfrac k8r(a^2-r^2)$