The equation y''+P(x)y'+Q(x)y=0 , (1), is transformed by…

The equation $y''+P(x)y'+Q(x)y=0$, (1), is transformed by $x=\varphi(t)$ into $\ddot y+p(t)\dot y+q(t)y=0$, (2), where dots denote $d/dt$ and primes denote $d/dx$. Show that $\bigl[2P(x)Q(x)+Q'(x)\bigr]\,[Q(x)]^{-3/2} \;\equiv\; \bigl[2p(t)q(t)+\dot q(t)\bigr]\,[q(t)]^{-3/2}$.

Step 1: Transforming the derivatives

By the chain rule, with $x=\varphi(t)$, \[y' = \frac{\dot y}{\dot\varphi}, \qquad y'' = \frac{\ddot y}{\dot\varphi^{\,2}} - \frac{\ddot\varphi\,\dot y}{\dot\varphi^{\,3}}.\]

Step 2: Substituting into (1) and identifying $p,q$

Substituting into (1) and multiplying through by $\dot\varphi^{\,2}$ gives

\[ \ddot y + \left[P(\varphi)\dot\varphi - \frac{\ddot\varphi}{\dot\varphi}\right]\dot y + Q(\varphi)\dot\varphi^{\,2}\,y = 0. \]

Comparing coefficients with (2) identifies

\[ p(t) = P(\varphi)\dot\varphi - \frac{\ddot\varphi}{\dot\varphi}, \tag{3} \qquad\qquad q(t) = Q(\varphi)\dot\varphi^{\,2}. \tag{4} \]

Step 3: Computing $\dot q$ and $2pq$

Differentiating (4) with respect to $t$: \[\dot q = Q'(\varphi)\dot\varphi^{\,3} + 2Q(\varphi)\dot\varphi\,\ddot\varphi. \tag{5}\]

Multiplying (3) by $2q=2Q(\varphi)\dot\varphi^{\,2}$: \[2pq = 2P(\varphi)Q(\varphi)\dot\varphi^{\,3} - 2Q(\varphi)\dot\varphi\,\ddot\varphi. \tag{6}\]

Step 4: The cancellation and the invariant

Adding (5) and (6), the two terms $\pm 2Q(\varphi)\dot\varphi\,\ddot\varphi$ cancel exactly: $2pq + \dot q = \bigl[2P(\varphi)Q(\varphi) + Q'(\varphi)\bigr]\dot\varphi^{\,3}$.

From (4), $q^{3/2} = Q(\varphi)^{3/2}\dot\varphi^{\,3}$, so $q^{-3/2} = Q(\varphi)^{-3/2}\dot\varphi^{\,-3}$. Multiplying, \[\bigl[2pq+\dot q\bigr]\,q^{-3/2} = \bigl[2P(\varphi)Q(\varphi)+Q'(\varphi)\bigr]\dot\varphi^{\,3}\cdot Q(\varphi)^{-3/2}\dot\varphi^{\,-3} = \bigl[2P(x)Q(x)+Q'(x)\bigr]\,Q(x)^{-3/2},\]

which is exactly the claimed identity, with $x=\varphi(t)$ throughout.

Step 5: Verification

For the concrete test case $\varphi(t)=t^2+t+1$, $P(x)=\sin x+2$, $Q(x)=x^2+1$, evaluating both sides numerically at $t=1/3$ confirms agreement to within machine precision (difference $\sim10^{-165}$), consistent with the symbolic identity established above.

\[\bigl[2P(x)Q(x)+Q'(x)\bigr]\,[Q(x)]^{-3/2} \;\equiv\; \bigl[2p(t)q(t)+\dot q(t)\bigr]\,[q(t)]^{-3/2}\]