Question

Using charpits method

(p+q)(z-px-qy)=1

Answer 1

$(p+q)(z-px-qy) =1\\ f=(p+q)(z-px-qy) -1=0$

Charpit's auxillairy equation is given as;

$\frac{dx}{-\frac{\partial f}{\partial p}}=\frac{dy}{-\frac{\partial f}{\partial q}}=\frac{dz}{-p\frac{\partial f}{\partial p}-q\frac{\partial f}{\partial p}}=\frac{dp}{\frac{\partial f}{\partial x}+p\frac{\partial f}{\partial z}}=\frac{dq}{\frac{\partial f}{\partial y}+q\frac{\partial f}{\partial z}}\\ \frac{\partial f}{\partial x}=-p^2-pq\\ \frac{\partial f}{\partial y}=-pq-q^2\\ \frac{\partial f}{\partial z}=p+q\\ \frac{\partial f}{\partial p}=z-xq-yq-2xp\\ \frac{\partial f}{\partial q}=z-xp-yp-2yq$

Now, the Charpit's equation is;

$\frac{dp}{0}=\frac{dq}{0}=\frac{dz}{2x(p^2+pq)+2y(q^2+pq)-2zp}=\frac{dy}{2q+xp+yp-z}=\frac{dx}{2xp+yq+xq-z}\\ \text{The first fraction implies, }\\ dp=0 \implies p=a\text{ By integrating }\\ \text{Similarly, the second fraction implies }\\ dq=0 \implies q=b\text{ By integrating }\\ \text{Now substitute into }dz=pdx+qdy\\ dz=adx+bdy\\ \text{Integrate through }\\ z=ax+by+c\\ \text{Substitute the value of z,p and q into f to get c}\\ (a+b)(ax+by+c-ax-by)=1\\ c=\frac{1}{a+b}\\ \implies z=ax+by+\frac{1}{a+b}\\ \implies (a+b)(z-ax-by) =1$