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.
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