Let y solve y'' = f(x) y , (*) , with f differentiable. Show…

Let $y$ solve $y'' = f(x)\,y$, $(*)$, with $f$ differentiable. Show that $z=y^2$ satisfies a third-order linear homogeneous differential equation, and find it.

Step 1: First and second derivatives of $z=y^2$

Differentiating $z=y^2$ twice by the product rule, $z' = 2yy', \qquad z'' = 2(y')^2 + 2yy''$.

Using $(*)$, $y''=fy$, so $yy''=fy^2$, and

\[ z'' = 2(y')^2 + 2fy^2. \]

Step 2: The third derivative

Differentiating $z''$ once more, treating $f=f(x)$: $z''' = 4y'y'' + 2f'y^2 + 4fyy'$.

Substituting $y''=fy$ into the first term, $4y'y''=4y'(fy)=4fyy'$, so the two $4fyy'$-type terms combine: $z''' = 4fyy' + 2f'y^2 + 4fyy' = 8fyy' + 2f'y^2$.

Step 3: Rewriting in terms of $z$ and $z'$

Since $z'=2yy'$, we have $yy'=\tfrac12 z'$, so $8fyy'=4fz'$; and $y^2=z$, so $2f'y^2=2f'z$. Hence

\[ z''' = 4f(x)\,z' + 2f'(x)\,z. \]

Every substitution used above was the linear relation $y''=fy$ — no nonlinear remainder survives — so this rearranges into a genuinely linear, homogeneous third-order equation for $z$:

\[ z''' - 4f(x)\,z' - 2f'(x)\,z = 0. \]

Step 4: Verification

Re-deriving from scratch, with $f$ kept as an abstract symbolic function: $z=y^2$ gives $z'=2yy'$; using $y''=fy$, $z''=2(y')^2+2fy^2$; differentiating again and substituting $y''=fy$ a second time, $z'''=8fyy'+2f'y^2$. Writing $yy'=z'/2$ and $y^2=z$ turns this into $z'''=4fz'+2f'z$, confirming $z'''-4fz'-2f'z=0$ identically in $f$ — the cancellation of the two $4fyy'$ contributions in Step 2 is exact and holds for arbitrary $f$, not just special choices.

\[z''' - 4f(x)\,z' - 2f'(x)\,z = 0\]