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I would like to solve the following equation $y^2=x^2+ax^2y^2+by^2x^3+cy^3x^2$ where $a,b,c$ are small, so $yapprox x+O(x^3)$. I would like to have a series approximation of the solution rather than an exact one, i.e. like $y=x+c_1x^3+c_2x^4+...$.

Is there a default function in mathematica that does that?

Quick-and-dirty solution:

y^2 - (x^2 + a x^2 y^2 + b y^2 x^3 + c y^3 x^2) /. 

       y -> x + c1 x^2 + c2 x^3 + c3 x^4 + O[x]^5 // FullSimplify

Solve[% == 0, {c1, c2, c3}]

y -> x + c1 x^2 + c2 x^3 + c3 x^4 + O[x]^5 /. % // TeXForm

$$yto x+frac{a x^3}{2}+frac{1}{2} (b+c) x^4+Oleft(x^5right)$$

Compare to the exact solution:

FullSimplify[

    Series[

        Solve[y^2 - (x^2 + a x^2 y^2 + b y^2 x^3 + c y^3 x^2) == 0, y][[3, 1, 2]]

    , {x, 0, 4}],

c > 0

]

Could use implicit differentiation, solve for all derivatives at x==0.

ee = y^2 - (x^2 + a x^2 y^2 + b y^2 x^3 + c y^3 x^2) /. y -> y[x];

dpolys = Table[D[ee, {x, j}], {j, 0, 5}] /. x -> 0



(* {y[0]^2, 2*y[0]*Derivative[1][y][0], -2 - 2*a*y[0]^2 - 2*c*y[0]^3 + 

     2*Derivative[1][y][0]^2 + 2*y[0]*Derivative[2][y][0], 

   -6*b*y[0]^2 - 12*a*y[0]*Derivative[1][y][0] - 

  18*c*y[0]^2*Derivative[1][y][0] + 

     6*Derivative[1][y][0]*Derivative[2][y][0] + 

  2*y[0]*Derivative[3][y][0], 

   -48*b*y[0]*Derivative[1][y][0] + 6*Derivative[2][y][0]^2 - 

     12*a*(2*Derivative[1][y][0]^2 + 2*y[0]*Derivative[2][y][0]) - 

     12*c*(6*y[0]*Derivative[1][y][0]^2 + 

     3*y[0]^2*Derivative[2][y][0]) + 

     8*Derivative[1][y][0]*Derivative[3][y][0] + 

  2*y[0]*Derivative[4][y][0], 

   -60*b*(2*Derivative[1][y][0]^2 + 2*y[0]*Derivative[2][y][0]) + 

     20*Derivative[2][y][0]*Derivative[3][y][0] - 

     20*a*(6*Derivative[1][y][0]*Derivative[2][y][0] + 

          2*y[0]*Derivative[3][y][0]) - 

  20*c*(6*Derivative[1][y][0]^3 + 

          18*y[0]*Derivative[1][y][0]*Derivative[2][y][0] + 

          3*y[0]^2*Derivative[3][y][0]) + 

  10*Derivative[1][y][0]*Derivative[4][y][0] + 

     2*y[0]*Derivative[5][y][0]} *)



soln = Solve[dpolys == 0]



(* Out[139]= {{y[0] -> 0, Derivative[1][y][0] -> -1, 

  Derivative[2][y][0] -> 0, 

     Derivative[3][y][0] -> -3*a, Derivative[4][y][0] -> 12*(-b + c)}, 

   {y[0] -> 0, Derivative[1][y][0] -> 1, Derivative[2][y][0] -> 0, 

     Derivative[3][y][0] -> 3*a, Derivative[4][y][0] -> 12*(b + c)}} *)

So two solutions, both getting up to the fourth order term. We create a table, then use the solutions to fill in values and recast as a Taylor polynomial for each solution branch.

derivs = Table[D[y[x], {x, j}], {j, 0, 4}] /. x -> 0;

derivs.(x^Range[0, 4]/Factorial[Range[0, 4]]) /. soln



(* Out[142]= {-x - (a x^3)/2 + 1/2 (-b + c) x^4, 

 x + (a x^3)/2 + 1/2 (b + c) x^4} *)

Substitute $y=xu$ to handle the node at the origin, differentiate and use AsymptoticDSolveValue:

x * AsymptoticDSolveValue[{

    D[y^2 == (x^2 + a x^2 y^2 + b y^2 x^3 + c y^3 x^2) /. 

       y -> x * u[x] // Simplify[#, x != 0] &, x],

    u[0] == 1}, u[x], {x, 0, 3}] // Collect[#, x] &

$$x+ frac{a x^3}{2}+frac{1}{2} x^4 (b+c)$$