Problem: Given the differential equation $$ x ^ 2 y & # 39; (x) + 4x , y (x) +2 , y (x) = r (x) $$ for a function $ r en C ^ 2 ( mathbb R) $and the initial values $$ y (0) = y_0, qquad y (0) = y_1, $$ how to analyze if a solution $ y in C ^ 2 ( mathbb R) $ exists locally around $ x_0 = $ 0 and / or is unique locally, depending on the choice of $ y_0 $ and $ y_1 $?
Some ideas
- Obviously, the Picard-Lindelöf theorem is not applicable around $ x_0 = $ 0 because the ODE is implicit. For $ x_0 neq $ 0however, we have an interval that does not contain $ 0 $, which gives an explicit ODE of the form $ y & # 39; + frac 4x y + frac {2} {x ^ 2} y = frac {r} {x ^ 2} $, where we observe that the homogeneous equation associated with the general solution $ y (x) = frac {c_1} {x} + frac {c_2} {x ^ 2} $ for $ c_1, c_2 in mathbb R $ be sure $ (0, + infty) $ or $ (- infty, 0) $.
- Yes $ c_1 neq 0 neq c_2 $then the solution is unlimited. However, this does not mean that the general solution of the inhomogeneous problem is not $ x_0 = $ 0 (which would have resulted in a contradiction with the existence of a solution).
- $ y (0) = y_0 $ implies that $ y_0 = frac {r (0)} {2} $.
- The first term of the ODE $ x ^ 2 , y & # 39; $ is differentiable with $ frac { mathrm} { mathrm dx} x ^ 2 y & # 39; (x) = lim_ {h to 0} frac {h ^ 2 y & # 39; (h) -0} {h} = lim_ {h to 0} h , y & # 39; (h) = $ 0 (using the continuity of $ y & # 39; $ and therefore the delimitation on a certain interval around $ 0 $). Although we do not know if $ y & # 39; & # 39; $ exists or not we still have an expression for the derivative of $ x ^ 2 y & # 39; $ at $ x_0 = $ 0. This means that after differentiating the ODE to $ x_0 = $ 0, we have $ 6 , y (0) = r & # 39; (x) $ or $ y_1 = frac {r ($)} {6} $.
- The last two points imply that ODE has no solution for $ y_0 neq r (0) / $ 2 and $ y_1 neq r $ (0) / $ 6. This reduces the problem to the case analysis $ y_0 = r (0) / $ 2, $ y_1 = r $ (0) / $ 6.
