Solve $p^2x+qy=z$

Question

Solve $ p^2x+qy=z$ where $p=z_x,\ q=z_y$ and $z=z(x,y)$.

I have tried it by Charpit's method.

Take $ f(x,y,z,p,q)=p^2x+qy-z=0$

Now $ f_x=p^2,\ f_y=q,\ f_z=-1,\ f_p=2px, f_q=y .$ So the Charpit's equation will be $$ \frac{dx}{2px}=\frac{dy}{y}=\frac{dz}{2p^2x+qy}=\frac{dp}{-(p^2-p)}=\frac{dq}{-(q-q)} $$ $$ \implies q=a\ \text{constant} $$ So, $p=\sqrt{\frac{z-ay}{x}}$ So, $$\sqrt{\frac{z-ay}{x}}dx +ady=dz$$ Now I stuck here. How to solve this?

Answer 1

Hint:

$xz_x^2+yz_y=z$

$z_x^2+2xz_xz_{xx}+yz_{xy}=z_x$

Let $u=z_x$ ,

Then $u^2+2xuu_x+yu_y=u$

$2xuu_x+yu_y=u(1-u)$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dy}{dt}=y$ , letting $y(0)=1$ , we have $y=e^t$

$\dfrac{du}{dt}=u(1-u)$ , we have $u=\dfrac{u_0e^t}{u_0e^t-1}=\dfrac{u_0y}{u_0y-1}$

$\dfrac{dx}{dt}=2xu=\dfrac{2xu_0e^t}{u_0e^t-1}$ , we have $x=(u_0e^t-1)^2f(u_0)=\dfrac{1}{(u-1)^2}f\left(\dfrac{u}{y(u-1)}\right)$ , i.e. $\dfrac{u}{u-1}=yF(x(u-1)^2)$

Answer 2

take $ay$ to RHS and then club $dz-ady= d(z-ay)$. $dx/x^{0.5}=d(z-ay)/(z-ay)^{0.5}$ > now it is solvable. $x^{0.5}=(z-ay)^{0.5}+b^{0.54}$