Problem source

Given that $y=x$ and $y=1-x^2$ satisfy  the differential equation
$$
\frac{\d^2 {y}}{\d x^2} + \p(x) \frac{\d {y}}{\d x} + \q(x) {y}=0\;,
\tag{*}
$$
show that $\p(x)= -2x(1+x^2)^{-1}$ and  $\q(x) = 2(1+x^2)^{-1}$. 
Show also that $ax+b(1-x^2)$ satisfies the differential equation for any constants $a$ and $b$. Given instead that $y=\cos^2(\frac{1}{2}x^2)$ and $y=\sin^2(\frac{1}{2}x^2)$ satisfy the  equation $(*)$, find $\p(x)$ and $\q(x)$.

Solution source

\begin{align*}
&& y &= x \\
&& y' &= 1 \\
&& y'' &= 0 \\
\Rightarrow && 0 &= 0 + p(x) + xq(x) \tag{1} \\
\\
&& y &= 1-x^2 \\
&& y' &= -2x \\
&& y'' &= -2 \\
\Rightarrow && 0 &= -2 -2x p(x)+(1-x^2)q(x) \tag{2}\\
\\
2x*(1) +(2): && 2 &= (2x^2+1-x^2) q(x) \\
\Rightarrow && q(x) &= 2(1+x^2)^{-1} \\
\Rightarrow && p(x) &= -2x(1+x^2)^{-1} \tag{by (1)}
\end{align*}

\begin{align*}
&& \frac{\d^2}{\d x^2} \left (a x + b(1-x^2) \right) + p(x) \frac{\d}{\d x} \left (a x + b(1-x^2) \right)+q(x) \left (a x + b(1-x^2) \right) \\
&&= a \frac{\d^2 x}{\d x^2} + b \frac{\d^2}{\d x^2} \left ( 1- x^2 \right) + ap(x) \frac{\d  x}{ \d x} + bp(x) \frac{\d }{\d x} \left ( 1- x^2 \right) + aq(x) x + bq(x)(1-x^2) \\
&&= a \left (\frac{\d^2 x}{\d x^2}+ p(x) \frac{\d  x}{ \d x} +q(x)x\right)+b \left ( \frac{\d^2}{\d x^2} \left ( 1- x^2 \right)+ p(x) \frac{\d }{\d x} \left ( 1- x^2 \right)+q(x)(1-x^2)\right) &= 0 
\end{align*}

\begin{align*}
&& y &= \cos^2(\tfrac12 x^2) = \frac12 \left (1 + \cos(x^2) \right) \\
&& y' &= -x \sin(x^2) \\
&& y'' &= -2x^2 \cos(x^2)-\sin(x^2) \\
\Rightarrow && 0 &= -2x^2 \cos(x^2)-\sin(x^2)+p(x)(-x \sin(x^2)) +\frac12 \left (1 + \cos(x^2) \right)q(x)  \\
\Rightarrow && 2x^2\cos(x^2)+\sin(x^2) &= -x \sin(x^2) p(x) + \frac12(1 + \cos(x^2)) q(x) \tag{3}\\ 
\\
&& y &= \sin^2(\tfrac12 x^2) = \frac12 \left ( 1 - \cos (x^2) \right) \\
&& y' &= x\sin(x^2) \\
&& y'' &= 2x^2 \cos(x^2)+\sin(x^2)  \\
\Rightarrow && 0 &=  2x^2 \cos(x^2)+\sin(x^2) +p(x) x \sin(x^2) + \frac12 \left ( 1 - \cos (x^2) \right)q(x)\\
\Rightarrow && -2x^2 \cos(x^2)-\sin(x^2) &= p(x) x \sin(x^2) + \frac12 \left ( 1 - \cos (x^2) \right)q(x) \tag{4}\\
(3)+(4): && 0 &= q(x) \\
\Rightarrow && p(x) &= -\frac{2x^2 \cos(x^2)+\sin(x^2)}{x \sin(x^2)}
\end{align*}