= Solution
Write a <power series> $y=\sum_{n\geq0}c_nx^n$. Substitution gives
$$
\sum_{n\geq0}n(n-2)c_nx^{n-1}+4\sum_{n\geq0}c_nx^{n+3}=0.
$$
The lowest coefficients give $c_1=0$, $c_3=0$, with $c_0,c_2$ free. For $n\geq4$, matching the coefficient of $x^{n-1}$ gives the recurrence
$$
\boxed{c_n=-\frac{4c_{n-4}}{n(n-2)}.}
$$
All odd coefficients consequently vanish. For the first solution, $c_0=a$, $c_2=y_1''(0)/2=0$, so the first three nonzero terms, when $a\ne0$, are
$$
\boxed{y_1(x)=a-\frac a2x^4+\frac a{24}x^8+O(x^{12}).}
$$
For the second, $c_0=0$, $c_2=b/2$, giving
$$
\boxed{y_2(x)=\frac b2x^2-\frac b{12}x^6+\frac b{240}x^{10}+O(x^{14}).}
$$
If 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 $\alpha\ne0$. The <chain rule> gives
$$
xy''-y'+4x^3y
=\alpha^2x^{2\alpha-1}Y_{uu}
+\alpha(\alpha-2)x^{\alpha-1}Y_u+4x^3Y.
$$
After 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> yields
$$
\boxed{\alpha=2,\qquad y(x)=A\cos(x^2)+B\sin(x^2).}
$$
It holds on each interval away from zero and both solutions extend smoothly through zero. Here $y(0)=A$ and $y''(0)=2B$, 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.
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