Show that the function y(x)=(arcsinsqrt x)/(√(x-x^2)) is a…
Show that the function \(y(x)=\dfrac{\arcsin\sqrt x}{\sqrt{x-x^2}}\) is a solution of the differential equation \(y'=f(x)y+\varphi(x)\), where \(f(x)\) and \(\varphi(x)\) are rational functions of \(x\) to be determined. Show that a particular solution of this equation can be expanded in a power series in a neighborhood of \(x=0\).
Explanation of Method
Compute \(y'\) directly with Sympy's \(\texttt{diff}\) (not the source's error-prone manual quotient rule) and confirm \(y'-f(x)y-\varphi(x)\) simplifies to exactly \(0\) for the claimed \(f(x)=\dfrac{2x-1}{2x(1-x)}\), \(\varphi(x)=\dfrac{1}{2x(1-x)}\) — settling this part cleanly, unlike the source's confused partial-fraction attempt in its own step 6 ("Wait, let me recalculate"). For the power-series part, the source assumed without checking that \(y\) is singular like \(x^{-1/2}\) at \(x=0\) — but Sympy's \(\texttt{limit}\) shows \(y(x)\to1\) as \(x\to0\) (since \(\arcsin\sqrt x\sim\sqrt x\) and \(\sqrt{x-x^2}\sim\sqrt x\) cancel), so \(y\) is actually regular at \(x=0\) even though both \(f\) and \(\varphi\) individually have simple poles there. The correct expansion is an ordinary Taylor series, computed directly with \(\texttt{series}\), and its coefficients are checked to satisfy \(y'=f y+\varphi\) order by order.
1. Compute \(y'\) directly and identify \(f,\varphi\)
Sympy differentiates \(y=\arcsin(\sqrt x)/\sqrt{x-x^2}\) directly and, after simplifying \(y'-\Big[\dfrac{2x-1}{2x(1-x)}y+\dfrac{1}{2x(1-x)}\Big]\), returns \(0\) identically. So the source's claimed \[ \boxed{f(x)=\frac{2x-1}{2x(1-x)}},\qquad \boxed{\varphi(x)=\frac{1}{2x(1-x)}} \] are correct — this repairs the source's step 6, whose partial-fraction manipulation visibly went wrong midway and was never actually needed to answer the question (Part 1 only asks for \(f,\varphi\), which are already in fully reduced rational form).
2. Determine the true behaviour of \(y\) at \(x=0\)
Sympy's \(\texttt{limit}(y,x,0)=1\): since \(\arcsin\sqrt x\sim\sqrt x-\tfrac16x^{3/2}+\cdots\) and \(\sqrt{x-x^2}=\sqrt x\sqrt{1-x}\sim\sqrt x\), the common factor \(\sqrt x\) cancels and \(y\) is finite and analytic at \(x=0\) — not singular like \(x^{-1/2}\), contrary to the source's step 7 assumption. \(f(x)\) and \(\varphi(x)\) each do have a simple pole \(\mp\frac{1}{2x}\) at \(x=0\), but these poles cancel exactly in the combination \(f(x)y+\varphi(x)\) because \(y(0)=1\): near \(x=0\), \(f\,y+\varphi\sim-\tfrac{1}{2x}\cdot1+\tfrac{1}{2x}=0\)+finite, matching a finite \(y'(0)\).
3. Compute the actual Taylor series with Sympy
\(\texttt{series}(y,x,0,6)\) gives
\[
y = 1+\tfrac23x+\tfrac{8}{15}x^2+\tfrac{16}{35}x^3+\tfrac{128}{315}x^4+\tfrac{256}{693}x^5+O(x^6).
\]
Substituting \(y=\sum a_nx^n\) into the exact (non-fractional) form of the ODE,
\(2x(1-x)y'-(2x-1)y=1\), and matching coefficients of \(x^n\) gives the two-term recursion
\((2n+1)a_n=2n\,a_{n-1}\) (\(n\ge1\)) with \(a_0=1\), whose closed form
\[
a_n=\frac{4^n(n!)^2}{(2n+1)!}
\]
is verified with Sympy to reproduce \(1,\ \tfrac23,\ \tfrac{8}{15},\ \tfrac{16}{35},\ \tfrac{128}{315},\
\tfrac{256}{693}\) for \(n=0,\dots,5\) exactly, matching the directly-computed series term for term. Since
\(a_n/a_{n-1}=2n/(2n+1)\to1\), the radius of convergence is \(1\), consistent with the domain
\(0 Truncating the series to \(O(x^6)\), Sympy computes both \(y_{\text{trunc}}'\) and
\(f\,y_{\text{trunc}}+\varphi\) as series in \(x\) up to \(O(x^5)\) (the constant \(1/(2x)\) poles in \(f,\varphi\)
cancelling exactly against the \(x^0\) term of \(y_{\text{trunc}}\), as anticipated in Step 2) and finds them
identical: \(2/3+16x/15+48x^2/35+512x^3/315+1280x^4/693+2048x^5/1001+O(x^6)\) on both sides. So the regular
power series genuinely satisfies \(y'=fy+\varphi\) to the order checked — confirming this is the correct
resolution, not the source's unexplained, never-executed pivot to a half-integer-power ansatz
\(y=\sum a_nx^{n+1/2}\) (which, as Step 2 shows, describes the singular behaviour of the homogeneous
solutions near the regular-singular point \(x=0\), not of the particular solution \(y(x)\) given in the
problem, which is analytic there). Final answer: Differential equation:
\[
\boxed{\ y' = \frac{2x-1}{2x(1-x)}\,y+\frac{1}{2x(1-x)}\ }
\]
confirmed by direct Sympy differentiation to hold identically for \(y=\arcsin\sqrt x/\sqrt{x-x^2}\). Power series at \(x=0\) (corrected: \(y\) is analytic at \(x=0\), not singular):
\[
\boxed{\ y(x)=\sum_{n=0}^\infty \frac{4^n(n!)^2}{(2n+1)!}\,x^n = 1+\frac23x+\frac{8}{15}x^2+\frac{16}{35}x^3+
\frac{128}{315}x^4+\frac{256}{693}x^5+\cdots,\qquad |x|<1\ }
\]
verified to satisfy the ODE order by order in Sympy.4. Confirm the series solves the ODE order by order
Result