Solving r²(dT/dr)+BrT=T² with substitution λ=T/r

[leomclaughlin]

(r^2) (dT/dr)+B*r*T=T^2, with initial condition dT/dr |_r=0 =0 where B is a constant


I've gotten it to this:

dT/dr = -BT/r + T2 / r2

by dividing everything by r2, then I substitute using λ= T/r which gives:


r * dλ/dr + lambda = -B * (λ) + λ^2


I don't know how to separate from here, any help is appreciated

 

[Strum]

Aren't you almost finished?

$r \frac{d\lambda}{dr} + \lambda = -B\lambda + \lambda^{2} \Rightarrow \frac{1}{-\lambda(B+1) + \lambda^{2}} \frac{d\lambda}{dr} = \frac{1}{r}$