Solution
The cylindrical and spherical coordinates are defined as given in the textbook figures shown below
The relation between these is given by (13) in the book\begin{align} z & =r\cos \theta \tag{1}\\ \rho & =r\sin \theta \tag{2}\\ \phi & =\phi \tag{3} \end{align}
To obtain the required formula, we will use the chain rule. Since in spherical we have u\equiv u\left ( r,\theta \right ) and in cylindrical we have u\equiv u\left ( \rho ,z\right ) , then by chain rule \frac{\partial u}{\partial \theta }=\frac{\partial u}{\partial \rho }\frac{\partial \rho }{\partial \theta }+\frac{\partial u}{\partial z}\frac{\partial z}{\partial \theta } But from (2) \frac{\partial \rho }{\partial \theta }=r\cos \theta and from (1) \frac{\partial z}{\partial \theta }=-r\sin \theta , hence the above becomes \frac{\partial u}{\partial \theta }=\frac{\partial u}{\partial \rho }\left ( r\cos \theta \right ) +\frac{\partial u}{\partial z}\left ( -r\sin \theta \right ) But r\cos \theta =z and -r\sin \theta =\rho , hence the above simplifies to\begin{equation} \frac{\partial u}{\partial \theta }=z\frac{\partial u}{\partial \rho }-\rho \frac{\partial u}{\partial z} \tag{4} \end{equation} Which is the result required to show. Now we need to show that \frac{\partial u}{\partial \theta } evaluated at boundary r=1,\theta =\frac{\pi }{2} is zero. But \theta =\frac{\pi }{2} implies that z=0, since z=r\cos \theta . Hence (4) now reduces to\begin{equation} \frac{\partial u}{\partial \theta }=-\rho \frac{\partial u}{\partial z} \tag{4} \end{equation} Since \theta =\frac{\pi }{2}, then \frac{\partial u}{\partial z} is the directional derivative normal to the base surface. But we are told it is insulated. This implies that \frac{\partial u}{\partial z}=0, since by definition this is what insulated means. Therefore \frac{\partial u}{\partial \theta }=0 at r=1,\theta =\frac{\pi }{2}, which is what we are asked to show.
Eq (6) in section 28 is y_{tt}\left ( x,t\right ) =a^{2}y_{xx}\left ( x,t\right ) -g At static displacement, by definition, there is no time dependency, hence y_{tt}=0 and the above becomes 0=a^{2}y_{xx}\left ( x,t\right ) -g Therefore now this becomes an ODE instead of a PDE since it does not depend on time, and we can write the above as\begin{equation} a^{2}y^{\prime \prime }\left ( x\right ) =g \tag{1} \end{equation} The boundary conditions y\left ( 0,t\right ) =0 and y\left ( 2x,t\right ) =0 now become y\left ( 0\right ) =0,y\left ( 2x\right ) =0. Now we need to solve (1) with these boundary conditions. This is an boundary value ODE. y^{\prime \prime }\left ( x\right ) =\frac{g}{a^{2}} The RHS is constant. The solution to the homogeneous ODE y^{\prime \prime }=0 is y_{h}=Ax+B. Let the particular solution be y_{p}=C_{3}x^{2}, then y_{p}^{\prime }=2C_{3}x and y_{p}^{\prime \prime }=2C_{3}. Substituting this in the above ODE gives\begin{align*} 2C_{3} & =\frac{g}{a^{2}}\\ C_{3} & =\frac{g}{2a^{2}} \end{align*}
Hence y_{p}\left ( x\right ) =\frac{g}{2a^{2}}x^{2}. Therefore the general solution is \begin{align} y & =y_{h}+y_{p}\nonumber \\ & =Ax+B+\frac{g}{2a^{2}}x^{2} \tag{2} \end{align}
Now we will use the boundary conditions to find A,B above. At x=0, (2) becomes 0=B Hence solution (2) reduces to\begin{equation} y\left ( x\right ) =Ax+\frac{g}{2a^{2}}x^{2} \tag{3} \end{equation} At x=2c, the second boundary condition gives\begin{align*} 0 & =2cA+\frac{g}{2a^{2}}\left ( 4c^{2}\right ) \\ A & =\frac{-g}{2a^{2}}\frac{\left ( 4c^{2}\right ) }{2c}\\ & =\frac{-gc}{a^{2}} \end{align*}
Hence the solution (3) becomes\begin{align} y & =\frac{-gc}{a^{2}}x+\frac{g}{2a^{2}}x^{2}\nonumber \\ y & =\frac{gx^{2}-2gcx}{2a^{2}} \tag{4} \end{align}
To get the result needed, we can manipulate this more as follows. From (4)\begin{align} 2a^{2}y & =gx^{2}-2gcx\nonumber \\ & =g\left ( x^{2}-2cx\right ) \nonumber \\ & =g\left ( x-c\right ) ^{2}-gc^{2}\nonumber \end{align}
Hence\begin{align*} g\left ( x-c\right ) ^{2} & =2a^{2}y+gc^{2}\\ \left ( x-c\right ) ^{2} & =\frac{2a^{2}y}{g}+c^{2}\\ & =\frac{2a^{2}}{g}\left ( y+\frac{gc^{2}}{2a^{2}}\right ) \end{align*}
Now since a^{2}=\frac{H}{\delta } then the above becomes\begin{align*} \frac{g}{2a^{2}}\left ( x-c\right ) ^{2} & =y+\frac{gc^{2}}{2a^{2}}\\ y & =\frac{1}{2a^{2}}\left ( g\left ( x-c\right ) ^{2}-gc^{2}\right ) \\ & =\frac{g}{2\frac{H}{\delta }}\left ( \left ( x-c\right ) ^{2}-c^{2}\right ) \\ & =\frac{\delta }{H}\frac{g}{2}\left ( \left ( x-c\right ) ^{2}-c^{2}\right ) \end{align*}
We see now that y is directly proportional to \delta and c^{2} and inversely proportional to H.
solution
The wave PDE in 1D is given by \begin{equation} y_{tt}\left ( x,t\right ) =a^{2}y_{xx}\left ( x,t\right ) \tag{1} \end{equation} Where a^{2}=\frac{H}{\delta } Where H is the tension in the strand and \delta is the mass per unit length of the strand. But weight=\left ( mass\right ) g. hence \delta =\frac{weight}{g}. We are given that weight=0.032 lb, and that g=32 ft/s^{2}. This implies that \delta =\frac{0.032}{32}=\frac{1}{1000} Hence a^{2}=\frac{10}{\frac{1}{1000}}=10^{4} Therefore (1) becomes\begin{equation} y_{tt}\left ( x,t\right ) =10^{4}y_{xx}\left ( x,t\right ) \tag{2} \end{equation} Since at t=0 we are told that strand lies along the x-axis, then y\left ( x,0\right ) =0 and problem says y_{t}\left ( x,0\right ) =1. For boundary conditions, since strand fixed at x=0 and x=1, then this implies y\left ( 0,t\right ) =0 and y\left ( 1,t\right ) =0. Therefore the PDE is\begin{align*} y_{tt}\left ( x,t\right ) & =10^{4}y_{xx}\left ( x,t\right ) \qquad 0<x<1,t>0\\ y\left ( x,0\right ) & =0\\ y_{t}\left ( x,0\right ) & =1\\ y\left ( 0,t\right ) & =0\\ y\left ( 1,t\right ) & =0 \end{align*}
Applying the first initial conditions y\left ( x,0\right ) =0 to the solution \begin{equation} y\left ( x,t\right ) =\phi \left ( x+at\right ) +\psi \left ( x-at\right ) \tag{1} \end{equation} Gives\begin{equation} 0=\phi \left ( x\right ) +\psi \left ( x\right ) \tag{2} \end{equation} But y_{t}=a\phi ^{\prime }-a\psi ^{\prime }. Hence the second initial conditions at t=0 gives\begin{equation} 0=a\phi ^{\prime }\left ( x\right ) -a\psi ^{\prime }\left ( x\right ) \tag{3} \end{equation} Taking derivative of (2) and multiplying the resulting equation by a gives\begin{equation} 0=a\phi ^{\prime }\left ( x\right ) +a\psi ^{\prime }\left ( x\right ) \tag{2A} \end{equation} Adding (3,2A) gives\begin{align*} 2a\phi ^{\prime }\left ( x\right ) & =0\\ \phi ^{\prime }\left ( x\right ) & =0 \end{align*}
Therefore \begin{equation} \phi \left ( x\right ) =C \tag{4} \end{equation} Where C is an arbitrary constant. Substituting the above result back in (2) gives\begin{align} 0 & =C+\psi \left ( x\right ) \nonumber \\ \psi \left ( x\right ) & =-C \tag{5} \end{align}
From (4,5) we see that \begin{align*} \phi \left ( x\right ) & =C\\ \psi \left ( x\right ) & =-C \end{align*}
Now applying boundary condition y\left ( 0,t\right ) =f\left ( t\right ) to (1) gives f\left ( t\right ) =\phi \left ( at\right ) +\psi \left ( -at\right ) But a is the speed of the wave given by a=\frac{x}{t} or t=\frac{x}{a}. Hence the above becomes\begin{align*} f\left ( \frac{x}{a}\right ) & =\phi \left ( x\right ) +\psi \left ( -x\right ) \\ \psi \left ( -x\right ) & =f\left ( \frac{x}{a}\right ) -\phi \left ( x\right ) \end{align*}
Since \phi \left ( x\right ) =C from equation (4), then the final result is obtained\begin{equation} \psi \left ( -x\right ) =f\left ( \frac{x}{a}\right ) -C\qquad x\geq 0 \tag{6} \end{equation}
Since the part to the right of x=at is unaffected by the movement of the right, then \begin{equation} y\left ( x,t\right ) =0\qquad x\geq at \tag{1} \end{equation} So now we need to find the solution for x<at and x\geq 0. From y\left ( x,t\right ) =\phi \left ( x+at\right ) +\psi \left ( x-at\right ) And using (6) in part (a), we see that \psi \left ( x-at\right ) =f\left ( \frac{-\left ( x-at\right ) }{a}\right ) -C. Therefore the above becomes y\left ( x,t\right ) =\phi \left ( x+at\right ) +f\left ( \frac{-\left ( x-at\right ) }{a}\right ) -C But also from part (a) \phi \left ( x+at\right ) =C. Hence the above simplifies to\begin{align} y\left ( x,t\right ) & =c+f\left ( \frac{-\left ( x-at\right ) }{a}\right ) -C\nonumber \\ & =f\left ( \frac{-x+at}{a}\right ) \nonumber \\ & =f\left ( t-\frac{x}{a}\right ) \qquad x<at \tag{2} \end{align}
Combining (1) and (2) shows that y\left ( x,t\right ) =\left \{ \begin{array} [c]{ccc}0 & & x\geq at\\ f\left ( t-\frac{x}{a}\right ) & & x<at \end{array} \right .
This requires just substitution of the function f\left ( t\right ) given into the solution found above which is \begin{equation} y\left ( x,t\right ) =\left \{ \begin{array} [c]{ccc}0 & & x\geq at\\ f\left ( t-\frac{x}{a}\right ) & & x<at \end{array} \right . \tag{1} \end{equation} But \begin{equation} f\left ( t\right ) =\left \{ \begin{array} [c]{ccc}\sin \pi t & & 0\leq t\leq 1\\ 0 & & t>1 \end{array} \right . \tag{2} \end{equation} Substituting (2) into (1) gives, after replacing each t in (2) by t-\frac{x}{a} the result needed y\left ( x,t\right ) =\left \{ \begin{array} [c]{ccc}0 & & x\geq at\\ \sin \left ( \pi \left ( t-\frac{x}{a}\right ) \right ) & & a\left ( t-1\right ) <x<at \end{array} \right .
We want to do the transformation from y\left ( x,t\right ) to y\left ( u,v\right ) . Therefore \frac{\partial y}{\partial x}=\frac{\partial y}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial y}{\partial v}\frac{\partial v}{\partial x} But \frac{\partial u}{\partial x}=1 and \frac{\partial v}{\partial x}=1, hence the above becomes \frac{\partial y}{\partial x}=\frac{\partial y}{\partial u}+\frac{\partial y}{\partial v} And\begin{align*} \frac{\partial ^{2}y}{\partial x^{2}} & =\frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial x}\right ) \\ & =\frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial u}+\frac{\partial y}{\partial v}\right ) \\ & =\frac{\partial }{\partial x}\frac{\partial y}{\partial u}+\frac{\partial }{\partial x}\frac{\partial y}{\partial v}\\ & =\left ( \frac{\partial ^{2}y}{\partial u^{2}}\frac{\partial u}{\partial x}+\frac{\partial ^{2}y}{\partial uv}\frac{\partial v}{\partial x}\right ) +\left ( \frac{\partial ^{2}y}{\partial v^{2}}\frac{\partial v}{\partial x}+\frac{\partial ^{2}y}{\partial vu}\frac{\partial u}{\partial x}\right ) \end{align*}
But \frac{\partial u}{\partial x}=1,\frac{\partial v}{\partial x}=1, hence the above becomes\begin{align} \frac{\partial ^{2}y}{\partial x^{2}} & =\frac{\partial ^{2}y}{\partial u^{2}}+2\frac{\partial ^{2}y}{\partial uv}+\frac{\partial ^{2}y}{\partial v^{2}}\nonumber \\ y_{xx} & =y_{uu}+y_{vv}+2y_{uv} \tag{1} \end{align}
Similarly, \frac{\partial y}{\partial t}=\frac{\partial y}{\partial u}\frac{\partial u}{\partial t}+\frac{\partial y}{\partial v}\frac{\partial v}{\partial t} But \frac{\partial u}{\partial t}=\alpha and \frac{\partial v}{\partial t}=\beta , hence the above becomes \frac{\partial y}{\partial t}=\alpha \frac{\partial y}{\partial u}+\beta \frac{\partial y}{\partial v} And
\begin{align*} \frac{\partial ^{2}y}{\partial t^{2}} & =\frac{\partial }{\partial t}\left ( \frac{\partial y}{\partial t}\right ) \\ & =\frac{\partial }{\partial t}\left ( \alpha \frac{\partial y}{\partial u}+\beta \frac{\partial y}{\partial v}\right ) \\ & =\alpha \frac{\partial }{\partial t}\left ( \frac{\partial y}{\partial u}\right ) +\beta \frac{\partial }{\partial t}\left ( \frac{\partial y}{\partial v}\right ) \\ & =\alpha \left ( \frac{\partial ^{2}y}{\partial u^{2}}\frac{\partial u}{\partial t}+\frac{\partial ^{2}y}{\partial uv}\frac{\partial v}{\partial t}\right ) +\beta \left ( \frac{\partial ^{2}y}{\partial v^{2}}\frac{\partial v}{\partial t}+\frac{\partial ^{2}y}{\partial uv}\frac{\partial u}{\partial t}\right ) \end{align*}
But \frac{\partial u}{\partial t}=\alpha and \frac{\partial v}{\partial t}=\beta , hence the above becomes\begin{align} \frac{\partial ^{2}y}{\partial t^{2}} & =\alpha \left ( \alpha \frac{\partial ^{2}y}{\partial u^{2}}+\beta \frac{\partial ^{2}y}{\partial uv}\right ) +\beta \left ( \beta \frac{\partial ^{2}y}{\partial v^{2}}+\alpha \frac{\partial ^{2}y}{\partial uv}\right ) \nonumber \\ & =\alpha ^{2}\frac{\partial ^{2}y}{\partial u^{2}}+\alpha \beta \frac{\partial ^{2}y}{\partial uv}+\beta ^{2}\frac{\partial ^{2}y}{\partial v^{2}}+\alpha \beta \frac{\partial ^{2}y}{\partial uv}\nonumber \\ y_{tt} & =\alpha ^{2}y_{uu}+\beta ^{2}y_{vv}+2\alpha \beta y_{uv} \tag{2} \end{align}
And to obtain y_{xt}, then starting from above result obtained \frac{\partial y}{\partial t}=\alpha \frac{\partial y}{\partial u}+\beta \frac{\partial y}{\partial v} Now taking partial derivative w.r.t. x gives\begin{align*} \frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial t}\right ) & =\frac{\partial }{\partial x}\left ( \alpha \frac{\partial y}{\partial u}+\beta \frac{\partial y}{\partial v}\right ) \\ & =\alpha \frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial u}\right ) +\beta \frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial v}\right ) \\ & =\alpha \left ( \frac{\partial ^{2}y}{\partial u^{2}}\frac{\partial u}{\partial x}+\frac{\partial ^{2}y}{\partial uv}\frac{\partial v}{\partial x}\right ) +\beta \left ( \frac{\partial ^{2}y}{\partial v^{2}}\frac{\partial v}{\partial x}+\frac{\partial ^{2}y}{\partial uv}\frac{\partial u}{\partial x}\right ) \end{align*}
But \frac{\partial u}{\partial x}=1,\frac{\partial v}{\partial x}=1, hence the above becomes\begin{align} \frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial t}\right ) & =\alpha \left ( \frac{\partial ^{2}y}{\partial u^{2}}+\frac{\partial ^{2}y}{\partial uv}\right ) +\beta \left ( \frac{\partial ^{2}y}{\partial v^{2}}+\frac{\partial ^{2}y}{\partial uv}\right ) \nonumber \\ y_{xt} & =\alpha y_{uu}+\left ( \alpha +\beta \right ) y_{vu}+\beta y_{vv} \tag{3} \end{align}
Substituting (1,2,3) into Ay_{xx}+By_{xt}+Cy_{tt}=0 results in A\left ( y_{uu}+y_{vv}+2y_{uv}\right ) +B\left ( \alpha y_{uu}+\left ( \alpha +\beta \right ) y_{vu}+\beta y_{vv}\right ) +C\left ( \alpha ^{2}y_{uu}+\beta ^{2}y_{vv}+2\alpha \beta y_{uv}\right ) =0 Or y_{uu}\left ( A+B\alpha +C\alpha ^{2}\right ) +y_{uv}\left ( 2A+B\left ( \alpha +\beta \right ) +2C\alpha \beta \right ) +y_{vv}\left ( A+B\beta +C\beta ^{2}\right ) =0
Looking at the term above for y_{uu} we see it is A+B\alpha +C\alpha ^{2} which has the root\begin{align*} \alpha & =-\frac{b}{2a}\pm \frac{1}{2a}\sqrt{b^{2}-4ac}\\ & =-\frac{B}{2C}\pm \frac{1}{2C}\sqrt{B^{2}-4AC} \end{align*}
Hence if we pick the root \alpha =\alpha _{0}=-\frac{B}{2C}+\frac{1}{2C}\sqrt{B^{2}-4AC} then the term y_{uu} vanishes. Similarly for the term multiplied by y_{vv} which is A+B\beta +C\beta ^{2}. The root is \beta =-\frac{B}{2C}\pm \frac{1}{2C}\sqrt{B^{2}-4AC} And if we pick \beta =\beta _{0}=-\frac{B}{2C}-\frac{1}{2C}\sqrt{B^{2}-4AC} then the term y_{vv} vanishes also in the PDE obtained in part (a), and now the PDE becomes y_{uv}\left ( 2A+B\left ( \alpha +\beta \right ) +2C\alpha \beta \right ) =0 Substituting the above selected roots \alpha _{0},\beta _{0} into the above in place of \alpha ,\beta since these are the values we picked, then the above becomes\begin{align*} y_{uv}\left ( 2A+B\left ( -\frac{B}{2C}+\frac{1}{2C}\sqrt{B^{2}-4AC}-\frac{B}{2C}-\frac{1}{2C}\sqrt{B^{2}-4AC}\right ) +2C\alpha \beta \right ) & =0\\ y_{uv}\left ( 2A-\frac{2B^{2}}{2C}+2C\alpha \beta \right ) & =0 \end{align*}
And again replacing \alpha \beta above with \alpha _{0},\beta _{0} results in\begin{align*} y_{uv}\left ( 2A-\frac{2B^{2}}{2C}+2C\left ( -\frac{B}{2C}+\frac{1}{2C}\sqrt{B^{2}-4AC}\right ) \left ( -\frac{B}{2C}-\frac{1}{2C}\sqrt{B^{2}-4AC}\right ) \right ) & =0\\ y_{uv}\left ( 2A-\frac{2B^{2}}{2C}+2C\left ( \frac{B^{2}}{4C^{2}}+\frac{1}{4C^{2}}\left ( B^{2}-4AC\right ) \right ) \right ) & =0\\ y_{uv}\left ( 2A-\frac{2B^{2}}{2C}+\frac{B^{2}}{2C}+\frac{1}{2C}\left ( B^{2}-4AC\right ) \right ) & =0\\ y_{uv}\left ( 2A-\frac{2B^{2}}{2C}+\frac{B^{2}}{2C}+\frac{B^{2}}{2C}-2A\right ) & =0\\ \frac{B^{2}}{2C}y_{uv} & =0 \end{align*}
Since B\neq 0,C\neq 0 then the above simplifies to y_{uv}=0
Since y_{uv}=0 Or \frac{\partial }{\partial v}\left ( \frac{\partial y}{\partial u}\right ) =0 The implies that \frac{\partial y}{\partial u}=\Phi \left ( u\right ) Integrating w.r.t. u gives y\left ( u,v\right ) =\int \Phi \left ( u\right ) du+\psi \left ( v\right ) Where \psi \left ( v\right ) is the constant of integration which is a function.
Let \int \Phi \left ( u\right ) du=\phi \left ( u\right ) then the above can be written as y\left ( u,v\right ) =\phi \left ( u\right ) +\psi \left ( v\right ) Or in terms of x,t, since u=x+\alpha t and v=x+\beta t the above solution becomes y\left ( x,t\right ) =\phi \left ( x+\alpha t\right ) +\psi \left ( x+\beta t\right ) Where \phi ,\psi are arbitrary functions twice differentiable. When \alpha =+a,\beta =-a, then the above becomes y\left ( x,t\right ) =\phi \left ( x+at\right ) +\psi \left ( x-at\right ) Which is the general solution (7) in section (30). QED
The differential equation in problem 2 is Ay_{xx}+By_{xt}+Cy_{tt}=0 We want to do the transformation from y\left ( x,t\right ) to y\left ( u,v\right ) with \begin{align*} u & =x\\ v & =\alpha x+\beta t \end{align*}
Now \frac{\partial y}{\partial x}=\frac{\partial y}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial y}{\partial v}\frac{\partial v}{\partial x} But \frac{\partial u}{\partial x}=1 and \frac{\partial v}{\partial x}=\alpha , hence the above becomes \frac{\partial y}{\partial x}=\frac{\partial y}{\partial u}+\alpha \frac{\partial y}{\partial v} And \frac{\partial y}{\partial t}=\frac{\partial y}{\partial u}\frac{\partial u}{\partial t}+\frac{\partial y}{\partial v}\frac{\partial v}{\partial t} But \frac{\partial u}{\partial t}=0 and \frac{\partial v}{\partial t}=\beta , hence the above becomes \frac{\partial y}{\partial t}=\beta \frac{\partial y}{\partial v} Therefore\begin{align} \frac{\partial ^{2}y}{\partial x^{2}} & =\frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial x}\right ) \nonumber \\ & =\frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial u}+\alpha \frac{\partial y}{\partial v}\right ) \nonumber \\ & =\frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial u}\right ) +\alpha \frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial v}\right ) \nonumber \\ & =\left ( \frac{\partial ^{2}y}{\partial u^{2}}\frac{\partial u}{\partial x}+\frac{\partial ^{2}y}{\partial uv}\frac{\partial v}{\partial x}\right ) +\alpha \left ( \frac{\partial ^{2}y}{\partial v^{2}}\frac{\partial v}{\partial x}+\frac{\partial ^{2}y}{\partial vu}\frac{\partial u}{\partial x}\right ) \nonumber \\ & =\left ( \frac{\partial ^{2}y}{\partial u^{2}}+\alpha \frac{\partial ^{2}y}{\partial uv}\right ) +\alpha \left ( \alpha \frac{\partial ^{2}y}{\partial v^{2}}+\frac{\partial ^{2}y}{\partial vu}\right ) \nonumber \\ & =\frac{\partial ^{2}y}{\partial u^{2}}+\alpha \frac{\partial ^{2}y}{\partial uv}+\alpha ^{2}\frac{\partial ^{2}y}{\partial v^{2}}+\alpha \frac{\partial ^{2}y}{\partial vu}\nonumber \\ y_{xx} & =y_{uu}+\alpha ^{2}y_{vv}+2\alpha y_{uv} \tag{1} \end{align}
Similarly,\begin{align} \frac{\partial ^{2}y}{\partial t^{2}} & =\frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial t}\right ) \nonumber \\ & =\frac{\partial }{\partial x}\left ( \beta \frac{\partial y}{\partial v}\right ) \nonumber \\ & =\beta \left ( \frac{\partial ^{2}y}{\partial v^{2}}\frac{\partial v}{\partial t}+\frac{\partial ^{2}y}{\partial vu}\frac{\partial u}{\partial t}\right ) \nonumber \\ & =\beta \left ( \beta \frac{\partial ^{2}y}{\partial v^{2}}\right ) \nonumber \\ y_{tt} & =\beta ^{2}y_{vv} \tag{2} \end{align}
And to obtain y_{xt}, then starting from above result obtained \frac{\partial y}{\partial t}=\beta \frac{\partial y}{\partial v} Now taking partial derivative w.r.t. x gives\begin{align} \frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial t}\right ) & =\frac{\partial }{\partial x}\left ( \beta \frac{\partial y}{\partial v}\right ) \nonumber \\ & =\beta \left ( \frac{\partial ^{2}y}{\partial v^{2}}\frac{\partial v}{\partial x}+\frac{\partial ^{2}y}{\partial vu}\frac{\partial u}{\partial x}\right ) \nonumber \\ & =\beta \left ( \alpha \frac{\partial ^{2}y}{\partial v^{2}}+\frac{\partial ^{2}y}{\partial vu}\right ) \nonumber \\ y_{xt} & =\alpha \beta y_{vv}+\beta y_{vu} \tag{3} \end{align}
Substituting (1,2,3) into Ay_{xx}+By_{xt}+Cy_{tt}=0 results in A\left ( y_{uu}+\alpha ^{2}y_{vv}+2\alpha y_{uv}\right ) +B\left ( \alpha \beta y_{vv}+\beta y_{vu}\right ) +C\left ( \beta ^{2}y_{vv}\right ) =0 Or\begin{equation} Ay_{uu}+y_{uv}\left ( 2A\alpha +B\beta \right ) +y_{vv}\left ( A\alpha ^{2}+B\alpha \beta +C\beta ^{2}\right ) =0 \tag{4} \end{equation} Which is what asked to show.
Setting \alpha =\frac{-B}{\sqrt{4AC-B^{2}}},\beta =\frac{2A}{\sqrt{4AC-B^{2}}} in (4) above results in\begin{align*} Ay_{uu}+y_{uv}\left ( 2A\left ( \frac{-B}{\sqrt{4AC-B^{2}}}\right ) +B\left ( \frac{2A}{\sqrt{4AC-B^{2}}}\right ) \right ) +y_{vv}\left ( A\alpha ^{2}+B\alpha \beta +C\beta ^{2}\right ) & =0\\ Ay_{uu}+y_{vv}\left ( A\alpha ^{2}+B\alpha \beta +C\beta ^{2}\right ) & =0 \end{align*}
And the above now becomes\begin{align*} Ay_{uu}+y_{vv}\left ( A\left ( \frac{-B}{\sqrt{4AC-B^{2}}}\right ) ^{2}+B\left ( \frac{-B}{\sqrt{4AC-B^{2}}}\right ) \left ( \frac{2A}{\sqrt{4AC-B^{2}}}\right ) +C\left ( \frac{2A}{\sqrt{4AC-B^{2}}}\right ) ^{2}\right ) & =0\\ Ay_{uu}+y_{vv}\left ( \frac{AB^{2}}{4AC-B^{2}}-\frac{2B^{2}A}{4AC-B^{2}}+\frac{4CA^{2}}{4AC-B^{2}}\right ) & =0\\ Ay_{uu}+y_{vv}\left ( \frac{AB^{2}-2B^{2}A+4CA^{2}}{4AC-B^{2}}\right ) & =0\\ Ay_{uu}+Ay_{vv}\left ( \frac{-B^{2}+4CA}{4AC-B^{2}}\right ) & =0\\ Ay_{uu}+Ay_{vv} & =0\\ A\left ( y_{uu}+y_{vv}\right ) & =0 \end{align*}
Therefore, since A\neq 0 the above becomes y_{uu}+y_{vv}=0
Setting \alpha =-B,\beta =2A in (4) above results in\begin{align*} Ay_{uu}+y_{uv}\left ( -2AB+2AB\right ) +y_{vv}\left ( AB^{2}-2B^{2}A+4CA^{2}\right ) & =0\\ Ay_{uu}+y_{vv}\left ( 4CA^{2}-B^{2}A\right ) & =0\\ Ay_{uu}-Ay_{vv}\left ( B^{2}-4CA\right ) & =0 \end{align*}
But B^{2}-4CA=0, therefore the above becomes y_{uu}=0