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I am trying to integrate the following first-order ordinary differential equation and confirming whether it is possible such that $t(r) = 0$ when $r = 0$.

$$\int \frac{dr}{r^2((a - r)^2 + b)} = \int c dt$$

From computationally integrating the above equation, I have obtained $t(r)$ as the following.

$$t(r) = -\frac{a \log(a^2 - 2 a r + b + r^2)}{c (a^2 + b)^2} - \frac{a^2}{c r (a^2 + b)^2} - \frac{b}{c r (a^2 + b)^2} + \frac{2 a \log(r)}{c (a^2 + b)^2} + \frac{a^2 \tan^{-1}(\frac{r - a}{\sqrt b})}{(\sqrt{b} c (a^2 + b)^2)} - \frac{\sqrt b \tan^{-1}(\frac{r - a}{\sqrt b})}{c (a^2 + b)^2} + k_1$$

And, clearly this doesn't satisfy the required condition of $t = 0$ when $r = 0$.

On a second note, what are the required steps to solve the above differential equation analytically?

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  • $\begingroup$ Do a definite integral instead. Integrate the LHS from $r(0)$ to $r(t)$ and the RHS from $0$ to $t$, and the resulting expression for $r(t)$ should automatically satisfy the initial conditions since plugging in $t=0$ into the integrals with results in $\int_a^a = 0$ on both sides. $\endgroup$
    – whpowell96
    Feb 20 at 16:19
  • $\begingroup$ @whpowell96 will that always be the case? like when i did it indefinitely and tried to eliminate constants by enforcing the conditions, i find that it fails because $t \to -\infty$ when $r \to 0$, but i am not sure if i did something wrong while integrating $\endgroup$
    – Arjo
    Feb 20 at 16:26
  • $\begingroup$ It should work if the integral converges, but it doesn't in this case. Note that the constant function $r(t) = 0$ is a solution to $r' = r^2((a-r)^2+b)$. This solution was assumed not to exist when you separated the equation. $\endgroup$
    – whpowell96
    Feb 20 at 16:28
  • $\begingroup$ ahh right! thanks $\endgroup$
    – Arjo
    Feb 20 at 16:29

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