Question

Solve z=px+qy+√1+p

Answer 1

Solution;

Assume by Clairauts equation;

$z=ax+by+c$

Be a solution of ;

$\frac{\delta z}{\delta x}=a$ ;$\frac{\delta z}{\delta y}=b$

$p=a$ ; $q=b$

By substitution;

$z=ax+by+\sqrt{1+a}$ ...(i)

Which is a complete solution.

To find the general solution;

Put b=f(a)

By substitution;

$z=ax+f(a)y+\sqrt{1+a}$ .....(ii)

Differentiate partially with respect to a;

$0=x+f'(a)y+\frac{1}{2\sqrt{1+a}}$ ....(iii)

Elimanite 'a' from (ii) and (iii) to get the general solution.

To find the singular solution;

Differentiate (i) w.r.t a;

$0=x+\frac{1}{2\sqrt{1+a}}$ ....(iv)

$x=-\frac{1}{2\sqrt{1+a}}$

Differentiate (i) w.r.t b;

$0=y$ .....(v)

By substitution;

$z=-\frac{a}{2\sqrt{1+a}}+\sqrt{1+a}$ ...(vi)

Eliminate 'a' and 'b' between (i) and (vi) to find a singular solution.