How to Show that $\frac{\partial^2 z}{\partial t^2}=\alpha$? Suppose that a function $z=f(x, y)$ has continuous second partial derivatives, and can form the differential equation $$a\frac{\partial^2 z}{\partial x^2}+2b\frac{\partial^2 z}{\partial x \partial y}+c\frac{\partial^2 z}{\partial y^2}=0$$
The variables $a, b, c \not= 0$ and $b^2-ac=0$. Show that if $x=\frac{s}{b}+\frac{a}{b}t$ and $y=t$, then $\frac{\partial^2 z}{\partial t^2}=\alpha$
First, I simplified the equation to be $az''+2bz'+cz=0$. There is only one variable so I can solve for the differential equation by substitution. But I don't see how solving this equation will lead me to the answer. So, how do I show that $\frac{\partial^2 z}{\partial t^2}=\alpha$ is true?
 A: I am not sure what the role of $s$ is, I assume that it is an arbitrary real number. It turns out that $\alpha = 0$, in general it is equal to the right hand side of the partial differential equation for $z$.
The goal is to differentiate $z(s/b +(a/b)t,t)$ over $t$. This is a composition, so we will use the chain rule. I'll denote the partial derivatives as $z_x$, $z_y$, etc. for brevity.
$$\frac{d}{dt} z(s/b +(a/b)t,t) = z_x(s/b+(a/b)t,t)(a/b) + z_y(s/b+(a/b)t,t).$$
$$\frac{d^2}{dt^2}z(s/b +(a/b)t,t) = z_{xx}(s/b+(a/b)t,t)(a/b)^2 + z_{xy}(s/b+(a/b)t,t)(a/b) + z_{yx}(s/b+(a/b)t,t)(a/b) + z_{yy}(s/b+(a/b)t,t).$$
Since the second order derivatives are continuous, we have $z_{xy} = z_{yx}$. Therefore, by multiplying both sides by $b^2$ we get
$$b^2\frac{d^2}{dt^2}z(s/b +(a/b)t,t) = a^2z_{xx}(s/b+(a/b)t,t) + 2abz_{xy}(s/b+(a/b)t,t) + b^2z_{yy}(s/b+(a/b)t,t).$$
Now, since $b^2=ac$, we find that the right-hand side has $a$ as a common factor, if we divide it out, then we get
$$\frac{b^2}{a}\frac{d^2}{dt^2}z(s/b +(a/b)t,t) = az_{xx}(s/b+(a/b)t,t) + 2bz_{xy}(s/b+(a/b)t,t) + cz_{yy}(s/b+(a/b)t,t).$$
The right-hand side is equal to 0 by the partial differential equation for $z$, therefore we get
$$\frac{d^2}{dt^2}z(s/b +(a/b)t,t)=0.$$
