I am stuck on the following problem that says:
Find the complete integral of $z=px + qy +pq$, where $p={ \partial z \over \partial x},q={ \partial z \over \partial y},$ and initial conditions are $z = y^2 , x=0$.
My Attempt:
The given equation is : $f(x,y,z,p,q)=px+qy+pq-z$. $z=y^2 , x=0$ So, Charpit's auxiliary equations are given by:
$$ds={dp \over 0}={dq \over 0}= {dz \over z+pq}={dx \over x+q}={dy \over y+p}$$
Now, from $$ {dp \over 0}={dq \over 0} $$ we get $p=c_1$, $c_1$ being arbitrary constants. I use the original PDE to get
$$ q={z-c_1x \over y+c_1}$$
Now, I have to use $$dz=pdx+qdy=c_1dx+{z-c_1x \over y+c_1}dy$$ we get $$z(x,y) - c_1x = c_2(y+c_1)$$ Now, I am stuck, because we need to apply the initial conditions, which will never be satisfied by the complete integral
Can someone help me out? Thanks in advance for your time.
$$xz_x+yz_y+z_xz_y-z=0 \tag 1$$ You have correctly found a set of solutions of the form : $$z(x,y)=c_1x+c_2y+c_1c_2 \tag 2$$ This set of solutions is commonly called a "complete integral". But this is not the general solution of the PDE. They are other solutions than $(2)$.
For example : $$z(x,y)=-xy$$ $z_x=-y$ and $z_y=-x$
$xz_x+yz_y+z_xz_y-z=x(-y)+y(-x)+(-y)(-x)-(-xy)=0$ which proves that $z=-xy$ is a solution of the PDE. $$ $$
The solution of the PDE which satisfies the condition $z(0,y)=y^2$ is : $$\boxed{z(x,y)=y^2+\frac{1}{16}x^2-\frac12 xy} \tag 3$$ $z_x=\frac18 x -\frac12 y$
$z_y=2y -\frac12 x$
$xz_x+yz_y+z_xz_y-z=x(\frac18 x -\frac12 y)+y(2y -\frac12 x)+(\frac18 x -\frac12 y)(2y -\frac12 x)-(y^2+\frac{1}{16}x^2-\frac12 xy)$
After simplification $=0$ which proves that $(3)$ is a solution of the PDE.