# How to find the power series solution of this ordinary differential equation using MMA [duplicate]

I already know that the solution of this differential equation $$y''(x) - x*y(x) = 0$$ can be expressed by the following power series:

$$y(x)=c0(1+\frac{x^{3}}{2\times3}+\frac{x^{6}}{2\times3\times5\times6}+\cdots\dfrac{x^{3n}}{2\times3\times5\times6\times\cdots\times(3n-1)3n}+\cdots)+c1(x+\frac{x^{4}}{3\times4}+\frac{x^{7}}{3\times4\times6\times7}+\cdots\dfrac{x^{3n+1}}{3\times4\times6\times7\times\cdots\times(3n)(3n+1)}+\cdots)=c0(1+\sum _{i=1}^\infty \frac{x^{3 i}}{\prod _{j=1}^i (3 j-1) (3 j)})+c1(x+\sum _{i=1}^\infty \frac{x^{3 i+1}}{\prod _{j=1}^i (3 j) (3 j+1)})$$

In the above formula, c0 and c1 are arbitrary constants.

I want to know how to use MMA to find the power series solution of the above ordinary differential equation and the power series solution of the following differential equation: $$y'(x)^2=\frac{9}{y(x)}-1$$

Other examples for testing:

DSolve[{y[x]*y''[x] == 1 + y'[x]^2, y[0] == 1, y'[0] == 0}, y[x], x]


I also want to get the infinite power series solution of this above differential equation, but I haven't solved this problem yet.

For the first one, expansion around zero

ode = y''[x] - x y[x] == 0
AsymptoticDSolveValue[ode, y[x], {x, 0, 10}]


For second one

ode = (y'[x])^2 == 9/y[x] - 1
AsymptoticDSolveValue[ode, y[x], {x, 0, 10}]


with warning:

AsymptoticDSolveValue::asdb There are multiple solution
branches for the equations, but AsymptoticDSolveValue will return only


one

See help on symptoticDSolveValue for more info. This is new function added in 11.3

• Thank you very much. But strictly speaking, your solution is only approximate, not infinite series, which is slightly unsatisfactory. – A little mouse on the pampas Feb 14 at 10:28
• For first ODE, DSolve gives exact solution AiryAi[x] C[1] + AiryBi[x] C[2]. If we then expand the two Airy functions around 0, we get series involving exact values of Γ'. Thus Series[AiryAi[x] ,{x, 0, 10{] begins 1/(3^(2/3) Gamma[2/3]) - x/(3^(1/3) Gamma[1/3]) + x^3/( 6 3^(2/3) Gamma[2/3])`. How does one reconcile that with the asymptotic series above? – murray Feb 14 at 15:26