Question

solve by Charpit's method: (p²+q²)y= q²​

Answer 1

Answer:

My attempt at solution

p2y=qz−q2=a...(I)p2y=qz−q2=a...(I)

This equation is of the form f1(x,p)=f2(y,q)f1(x,p)=f2(y,q).

Its solution is given by dz=pdx+qdydz=pdx+qdy, upon integrating this we get value of zz.

From (I)

−yq2+zq−a=0−yq2+zq−a=0, solving the quadratic equation for qq, we get

q=−z±z2−4ay√−2yq=−z±z2−4ay−2y.

Taking the positive value only, q=−z+z2−4ay√−2yq=−z+z2−4ay−2y .

Also, from (I), p2y