Discussion
The integral surface satisfying the partial differential equation \(\frac{\partial z}{\partial x} + z^{2}\frac{\partial z}{\partial y} = 0\) and passing through the straight line \(x = 1, y = z\) is
Answer: D
The integral surface is \((x - 1)z^{2} + z = y\).
This is a quasi-linear equation \(Pp + Qq = R\) with \(P = 1\), \(Q = z^{2}\), \(R = 0\). Lagrange's auxiliary equations are \(\frac{dx}{1} = \frac{dy}{z^{2}} = \frac{dz}{0}\), giving the characteristics:
\[ \begin{aligned} z &= c_{1} \\ y - z^{2}x &= c_{2} \end{aligned} \]
On the initial line write \(x = 1\), \(y = z = t\). Then \(c_{1} = t\) and \(c_{2} = t - t^{2}\), so \(c_{2} = c_{1} - c_{1}^{2}\). Substituting back:
\[ y - z^{2}x = z - z^{2} \;\Rightarrow\; (x - 1)z^{2} + z = y. \]
Check: at \(x = 1\) it reduces to \(z = y\). All four surfaces happen to contain that line, so the equation decides. On the other three, \(z\) is not constant along the characteristics; for example on \(x^{2} + y^{2} - z^{2} = 1\), \(z_{x} = x/z\) and \(z_{y} = y/z\), so \(z_{x} + z^{2}z_{y} = x/z + yz \ne 0\).
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