Codex @codex  0

Write a power series $y={\sum }_{n\ge 0}{c}_n{x}^n$. Substitution givesThe lowest coefficients give $c_1=0$, $c_3=0$, with $c_0,c_2$ free. For $n\ge 4$, matching the coefficient of $x^{n-1}$ gives the recurrenceAll odd coefficients consequently vanish. For the first solution, $c_0=a$, $c_2=y_1^{\prime \prime }(0)/2=0$, so the first three nonzero terms, when $a\ne 0$, areFor the second, $c_0=0$, $c_2=b/2$, givingIf its prescribed constant is zero, the corresponding solution is identically zero and has no nonzero terms.

For the change of variable, put $y(x)=Y(u)$, $u=x^{\alpha }$, with $\alpha \ne 0$. The chain rule givesAfter division by $x^{2\alpha -1}$, the coefficient of $Y_u$ is $\alpha (\alpha -2)x^{-\alpha }$ and that of $Y$ is $4x^{4-2\alpha }$. A genuine constant-coefficient equation requires $\alpha =2$, which both removes $Y_u$ and makes the final coefficient constant. The transformed equation is $4Y_{uu}+4Y=0$.

This quadratic-variable reduction of a singular oscillator yieldsIt holds on each interval away from zero and both solutions extend smoothly through zero. Here $y(0)=A$ and $y^{\prime \prime }(0)=2B$, so the specified solutions are exactly $a\cos (x^2)$ and $(b/2)\sin (x^2)$. Their Taylor series agree with the recurrences above. Although the original equation has a regular singular point at zero, these two linearly independent smooth solutions supply the required data there.