The heat conduction equation in two space dimensions may be expressed in terms of polar coordinates as $$k(u_{rr} +\dfrac{1}{r} u_r +\dfrac{1}{r^2} u_{θθ}) = u_t.$$ Assuming that $u(r, θ, t) = R(r) \Theta(θ)T (t)$, find ordinary differential equations satisfied by $R(r), \Theta(θ)$, and $T (t)$.
This is question question 49 on page 215 of Vector Calculus by Marsden and Tromba. The given answers are:
$T' + kc_1T = 0, , \Theta'' +c_2\Theta = 0, r^2R' +r R' −c_3R = 0$ for some constants $c_1, c_2$, and $c_3$.
I have managed to get the first two, but not the last equation. I started by writing $$k(R'' \Theta T + \dfrac{1}{r} R' \Theta T + \dfrac{1}{r^2} R \Theta'' T) = R \Theta T'.$$
I then tried to separate variables, to get one side depending on $t$ only and another depending on $(r,\theta)$. This would mean that both sides would be equal to a common constant, according to examples I have seen. This means that I get $$ \dfrac{R'' \Theta + \dfrac{1}{r} R' \Theta + \dfrac{1}{r^2} R \Theta''}{R\Theta} = \dfrac{R''}{R} + \dfrac{1}{r} \dfrac{R'}{R} + \dfrac{1}{r^2} \dfrac{\Theta''}{\Theta} = \dfrac{T'}{kT} = c_1.$$
This gave me the first equation, which is $T' = kc_1 T$. I then get the other equation as $\dfrac{R''}{R} + \dfrac{1}{r} \dfrac{R'}{R} + \dfrac{1}{r^2} \dfrac{\Theta''}{\Theta} = c_1$ or $$r^2 c_1 - r^2 \dfrac{R''}{R} - r\dfrac{R'}{R} = \dfrac{\Theta''}{\Theta} = c_2,$$ so I obtain $\Theta'' = c_2 \Theta$ and $r^2c_1 - r^2 \dfrac{R''}{R} - r\dfrac{R'}{R} = c_2$ which gave me $$r^2 R'' + rR' + c_2R - r^2 Rc_1 = 0. $$ This is somewhat similar to the printed answer, but I got an extra term with $r^2$. Please, could someone let me know where I went wrong? Thank you very much.