I have the following question:
A complete integral of $f=xpq+yp^2-1=0$ obtained by Charpit's method is:
$(a) \ (z+b)^2=4(ax-y)$
$(b) \ (z+b)^2=4(ax+y)$
$(c) \ (z-b)^2=4(ax+y)$
$(d) \ (z+b)^2=4ax+y^2$
where $p=z_x,q=z_y$.
The given answer is $(b)$.
I tried solving this by Charpit's method but couldn't get an answer. So I decided to see if the given answer satisfies the original equation.
Differentiating $(b)$ with respect to $x$: $$2(z+b)p=4a$$ Differentiating $(b)$ with respect to $y$: $$2(z+b)q=4$$
Substituting this in the original equation: $$\frac{4ax}{(z+b)^2}+\frac{4a^2y}{(z+b)^2}-1\neq0$$
Now, I tried doing the same for all other options and saw that none of them satisfy the original equation.
Am I missing something here?
If the equation were $f=xpq+yq^2−1=0$ the Charpit equations would be ($f_z=0$) \begin{array}{cccccccccc} &dx&:&dy&:&dz&:&-dp&:&-dq\\ =&f_p&:&f_q&:&f_pp+f_qq&:&f_x&:&f_y \\ =&xq&:&xp+2yq&:&2&:&pq&:&q^2 \end{array} For $q\ne 0$ the last two components integrate to $p=aq$.
The first and the last give $dx:x=-dq:q$ integrating to $qx=c$. Now insert into the equation $$ 1=axq^2+yq^2\iff \frac{x^2}{c^2}=ax+y $$ and $$ dz=pdx+qdy=qd(ax+y)=qd\left(\frac{x^2}{c^2}\right)=\frac{2}{c}dx $$ so that after integration $$ z+b=2\frac{x}{c}\implies (z+b)^2=4\frac{x^2}{c^2}=4(ax+y) $$ which is compatible with variants (b) and (c).
With the equation as given you get $p^2-q^2=a$ as one invariant of the characteristics, and then $z+b=\frac{2q}{ap}$ and with the original equation $$ a\left(\frac a2x(z+b)+y\right)=1-\frac{a^2}4(z+b)^2 \\\iff a^2(z+b+x)^2-a^2x^2+4y=4 $$ which does not look like any of the variants