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Substitute the power series and its termwise derivatives into the ordinary differential equation. The coefficient of $x^n$, for $n\ge 0$, isThe bracket factors as $(n-3)(n-4)$, so the power-series solution of a differential equation satisfiesThe odd and even recurrence relations terminate: $a_3=-a_1$, $a_5=0$, and $a_2=-6a_0$, $a_4=a_0$, $a_6=0$. Thus the full solution is $y=a_0(1-6x^2+x^4)+a_1(x-x^3)$. The two polynomials have independent initial data at zero, and the leading coefficient $1+x^2$ never vanishes on the real axis, so every real solution of a differential equation for this equation is their linear combination on the entire real axis. The given initial data select $a_0=0$, $a_1=1$:

With $t=1-x^2$ the chain rule gives $y^{\prime }=-2x{u}^{\prime }$ and $y^{\prime \prime }=-2u^{\prime }+4x^2{u}^{\prime \prime }$. Substitution, initially for $x>0$, givesThe transformed coefficients are $P(t)=1/t$ and $Q(t)=1/[4t(1-t)]$. At $t=0$, both $tP$ and $t^{2}Q$ are analytic; at $t=1$, both $(t-1)P$ and $(t-1{)}^{2}Q$ are analytic. Each point is singular because a normalized coefficient has a pole, but satisfies the regular singular point criterion for a second-order equation. Thus both points are regular singular points.

For a Frobenius method substitution $u=t^{\rho }{\sum }_{n\ge 0}{a}_n{t}^n$, the lowest power gives the indicial equationThe repeated indicial root is zero. Setting $\rho =0$ and equating the coefficient of $t^n$ givesConsequently the analytic power-series solution of a differential equation isTaking $a_0=2\pi$ reproduces the ellipse perimeter expansion.

For a second independent solution, normalize $u_1(0)=1$. Reduction of order and $P=1/t$ giveSince $u_1(s{)}^{-2}=1+O(s)$, this has the Logarithmic Frobenius solution formThe logarithm is $\log |t|$ on the real negative side, or a chosen complex logarithm on a slit complex neighborhood. Its nonzero logarithmic coefficient establishes independence. The physical perimeter is finite at $t=0$, so it selects the analytic branch rather than this logarithmic one.

Use a power-series solution of a differential equation, $y={\sum }_{n\ge 0}{c}_n{x}^n$. Comparing coefficients givesIn particular $c_0=0$, while $c_1$ and $c_2$ are free. The two prescribed derivative pairs select $(c_1,c_2)=(a,0)$ and $(0,b)$. The next coefficients are $c_3=c_1/2$, $c_4=c_2/6$, $c_5=c_1/24$, $c_6=c_2/120$. HenceThese are the first three nonzero terms when the respective parameter is nonzero; if $a=0$ or $b=0$, the corresponding solution vanishes identically.

For closed forms, set $y=xu$. Cancellation of the first-derivative terms reduces the equation, for $x\ne 0$, to $u^{\prime \prime }-u=0$. The analytic solutions extend across zero, and their hyperbolic cosine and hyperbolic sine expansions identifyThis is the hyperbolic reduction of a regular-singular differential equation: although the original leading coefficient vanishes at zero, both selected power-series solutions of a differential equation are entire.

For a normalized equation$x_0$ is an ordinary point when $P,Q$ are analytic there. It is a regular singular point when $(x-x_0)P(x)$ and $(x-x_0{)}^{2}Q(x)$ are analytic there. These are the ordinary point criterion for a second-order equation and regular singular point criterion for a second-order equation.

Hereso $x=0$ is regular singular. Seek a power-series solution of a differential equationEquating the coefficient of $x^n$ givesand henceThereforeThe series terminates exactly when $\lambda =N$ is a nonnegative integer: the factor with $j=N$ then makes $a_{N+1}=0$. Thus the polynomial solutions occur for