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For an invertible fundamental matrix, differentiating $Y_{\lambda }=AY$ in $z$ and $Y_z=BY$ in $\lambda$ gives $Y_{\lambda z}=(A_z+AB)Y$ and $Y_{z\lambda }=(B_{\lambda }+BA)Y$. Thus the necessary and sufficient compatibility identity isHere both derivatives must use the actual deformation variable $z$; the printed $t$ has not been defined. The isomonodromic deformation identity can be checked as a polynomial in $\lambda$, without assuming the conclusion.

First retain the coefficients actually printed in the PDF. Write $p=u^{\prime }$, $d=w+z/2$, and introduce the two residualsDirect matrix multiplication and differentiation giveFor example, the quadratic lower entry is the difference between $[D,B_0{]}_{21}=2w/u$ and the contribution $(A_1{)}_{21}=-zw/u$; it does not cancel. The constant lower entry follows by differentiating $2u^{\prime }w/u^2$ and collecting it as $2wH/u^2+2u^{\prime }K/u^2$. These entries also provide independent checks of the two most important cancellations.

Work on an open connected deformation domain where $u\ne 0$, as the matrices require. The quadratic entry forces $(2-z)w=0$ throughout the domain, hence $w\equiv 0$ by continuity, including a possible point $z=2$. All remaining entries then vanish exactly when $u^{\prime \prime }+(z/2)u=0$. Thus the literal answer isThis is a rescaled Airy function equation: with $\xi =-2^{-1/3}z$, the equation is $u_{\xi \xi }-\xi u=0$, so $u=c_1\mathrm{Ai}(\xi )+c_2\mathrm{Bi}(\xi )$ on a domain avoiding its zeros. In particular, choose the local solution with $u(1)=u^{\prime }(1)=1$ and $w=0$. It gives compatible matrices but violates the printed $w=u^{\prime }$. This proves that the requested assertion is false for the original coefficients, independently of the undefined $t/z$ in the printed differential equation.

For completeness, use the explicitly corrected lower coefficient $(A_1{)}_{21}=-2w/u$ from the preceding solution. The same multiplication gives the much simpler residualConsequently compatibility is equivalent to $K=H=0$. The first equation integrates without dividing by $w$:The constant of integration $C$ therefore satisfies $w=C{u}^2$, and the second equation becomesConversely these two equations give $K=H=0$, and substitution makes every residual entry vanish, proving sufficiency as well as necessity. This is the quadratic spectral Lax pair for zero-parameter Painlevé II. For $C\ne 0$, choose constants $b,a$ with $b^3=-2$ and $C{a}^2{b}^2=-2$, set $z=bx$, $u=aq(x)$, and obtain $q^{\prime \prime }=2q^3+xq$, the zero-parameter Painlevé II equation. For $C=0$ it reduces to the Airy function equation. Even with this correction, the conclusions are $w=C{u}^2$ and the coefficient $z/2$, rather than the incompatible expressions printed in the question.