Answer to Question #226713 in Mechanical Engineering for Sai

Question #226713
Solve z=px+qy+√1+p
1
Expert's answer
2021-08-18T03:51:01-0400

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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