If $p^2=a^2(\cos x)^2+b^2(\sin x)^2$, prove that $p +\frac{d^2 p}{dx^2}=\frac{a^2 b^2}{p^3}$ How to prove this? It seems simple enough to start but the at the end I cannot prove the expression $p +\frac{d^2 p}{dx^2}=\frac{a^2 b^2}{p^3}$. Is there a particular trick I am missing or is this just an ordinary sum with lots of manipulation. I did try modifying the question slightly by multiplying both the LHS and RHS by $p^3$. That seems to lessen the manipulation somewhat but still the answer does not come.
 A: $$p^2 = a^2 \cos^2 x + b^2(1 -\cos ^2 x) = (a^2-b^2)\cos^2x +b^2 = (a^2-b^2)\left(\frac{1}{2} + \frac{\cos 2x}{2} \right) +b^2 \\= \frac{a^2+b^2}{2} + \frac{(a^2-b^2)\cos 2x}{2}$$
So differentiating, we get
$$2pp' = -(a^2-b^2)\sin 2x \tag{1}$$
$$2(p')^2 + 2pp'' = -2(a^2-b^2)\cos 2x \tag{2} $$
Adding $(2) \times p^2$ and  $-\frac{1}{2}(1)^2$, we get
$$-2(pp')^2 + 2(pp')^2 + 2p^3p'' =- \frac{(a^2-b^2)^2\sin^2 2x}{2}   -2(a^2-b^2)\cos 2x \cdot p^2$$
Now adding $2p^4$ to both sides,
\begin{align}
2p^4 + 2p^3p'' &= -\frac{(a^2-b^2)^2\sin^2 2x}{2}   -2p^2[(a^2-b^2)\cos 2x \cdot- p^2] \\
&= -\frac{(a^2-b^2)^2\sin^2 2x}{2} -2p^2\left( \frac{(a^2-b^2)\cos 2x}{2} - \frac{a^2+b^2}{2} \right)\\
&=- \frac{(a^2-b^2)^2\sin^2 2x}{2}\\ 
&\ \ -2 \left(\frac{a^2+b^2}{2} + \frac{(a^2-b^2)\cos 2x}{2} \right) \left( \frac{(a^2-b^2)\cos 2x}{2} - \frac{a^2+b^2}{2} \right)\\
&=- \frac{(a^2-b^2)^2\sin^2 2x}{2} - 2\left( \left(\frac{(a^2-b^2)\cos 2x}{2} \right)^2-\frac{(a^2+b^2)^2}{4}    \right)\\
&= \frac{-(a^2-b^2)^2\sin^2 2x -(a^2-b^2)^2\cos^2 2x + (a^2 +b^2)^2 }{2}\\
&==\frac{(a^2+b^2)^2 - (a^2-b^2)^2}{2} = 2a^2b^2 
\end{align}
That is,
$$p^4 + p^3p'' = a^2b^2 $$
