Question

The complete integral of
$(z-px-qy{)}^3=pq+2(p^2+q{)}^2$ is

• A. $z=ax+by+\sqrt[3]{3pq+2(p^2+q{)}^2}$
• B. $z=ax+by+\sqrt[3]{3ab+2(a^2+b{)}^2}$
• C. $z=ax+by+\sqrt[3]{3ab}+\sqrt[3]{32(a^2+b{)}^2}$
• D. $z=ax+by+c$

Answer 1

The correct option is B $z=ax+by+\sqrt[3]{3ab+2(a^2+b{)}^2}$

General form of Clairaut's equation is
$z=px+qy+pq$ .... (i)
and it's solutioin is
$z=ax+by+ab$ ..... (ii)
where $p=\frac{\partial z}{\partial x}$ and $q=\frac{\partial z}{\partial y}$ and $a,b$ are arbitrary constant.
Using $2$ we can easily get $\frac{\partial z}{\partial x}=a$ and $\frac{\partial z}{\partial y}=b$
Hence to find general solution of Clairaut equation, we can replace $p$ by $a$ and $q$ by $b$.
Now, given equation is:
$(z-px-qy{)}^3=pq+2(p^2+q{)}^2$
$z=px+qy+[pq+2(p^2+q){]}^{1/3}$
So complete solution is
$z=ax+by+[ab+2(a^2+b){]}^{1/3}$