How to solve this integral of an ODE?

Question

https://www.symbolab.com/solver/integral-calculator/%5Cint%5Cint%5Cleft(Ax%5E%7B2%7D%2B%5Cfrac%7BB%7D%7Bx%7D%2B%5Cfrac%7B2%7D%7Bx%5E%7B2%7D%7D%5Cright)dxdx

I have used this symbolic solver to solve for this ODE but still couldnt find the answer. Please provide proof of your answers below

Answer 1

Consider the equation $\left(\frac{d^2}{dr^2} - \frac{2}{r^2}\right) y = y'' - \frac{2}{r^2}y = 0.$ This can be recognized as a Cauchy-Euler equation, so we can look for solutions of the form $y = r^p.$ This gives us: $$p(p-1)r^{p-2} - 2r^{p-2} = 0 \Rightarrow r^{p-2}\left(p^2 - p - 2\right) = 0.$$

Now given that $r$ is not $0,$ we can divide both sides by $r^{p-2}$ to obtain $p^2 - p - 2 = 0,$ which has the solutions $p = -1, p = 2.$ This gives us the two solutions $y_1 = r^{-1}, y_2 = r^2.$ We can verify that these are linearly independent by taking their Wronskian: $det(W) = r^{-1}(2r) - r^2(-r^{-2}) = 3 \neq 0,$ so we have a fundamental set of solutions and the general solution is $y = c_1r^{-1} + c_2r^2.$

Now we know that any solution $f$ to the original equation must satisfy $f'' - \frac{2}{r^2}f = c_1r^{-1} + c_2r^2,$ and likewise every solution to this equation will solve the original. Using our results from before we have that the complementary solution is $f_c = k_1r^{-1} + k_2r^2$ for any arbitrary constants $k_1, k_2.$

For the particular solution we can use the method of variation of parameters: assume the form $f_p = u_1y_1 + u_2y_2.$ Skipping the derivation (which can be found here if you're unfamiliar)

$$u_1' = \frac{-r^2(c_1r^{-1}+c_2r^2)}{3} = -\frac{c_1}3 r - \frac{c_2}3 r^4 \Rightarrow u_1 = -\frac{c_1}{6} r^2 -\frac{c_2}{15}r^5$$

$$u_2' = \frac{r^{-1}(c_1r^{-1}+c_2r^2)}{3} = \frac{c_1}{3}r^{-2} + \frac{c_2}{3}r \Rightarrow u_2 = -\frac{c_1}{3}r^{-1} + \frac{c_2}{6}r^2$$

Plugging into our form for $f_p$ gives us

$$f_p = (-\frac{c_1}{6} r^2 -\frac{c_2}{15}r^5)r^{-1} + (-\frac{c_1}{3}r^{-1} + \frac{c_2}{6}r^2)r^2 = -\frac{c_1}2 r + \frac{c_2}{10} r^4$$

Letting $k_3 = -\frac{c_1}2, k_4 = \frac{c_2}{10}$ and adding this to our complementary solution, we get $f(r) = k_1r^{-1} + k_2r^2 + k_3r + k_4 r^4,$ which is our general solution for the original equation.

Hope this helps!

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Quick note: there is almost surely a more straight-forward way of solving this by first squaring the operator and then solving as one big equation, but I chose this method for two main reasons: first, it's not too difficult to carry out, whereas I think the other method would get a bit messier, and second, I just thought this idea was a bit neat.