Answer to Question #164573 in Differential Equations for Ramya

Question

Find the integral surface of the equation pq=z passing through C : x₀ = 0, y₀ = s, z₀ = s²

Answer 1

We use the Charpit's method for solution.

$f(x,y,z,p,q)=pq-z=0$ with $p=z_x$ and $q=z_y$.

Charpit's auxiliary equations in general case:

$\frac{-dx}{\frac{\partial f}{\partial p}}= \frac{-dy}{\frac{\partial f}{\partial q}}= \frac{-dz}{p\frac{\partial f}{\partial p} +q\frac{\partial f}{\partial q}}=\frac{dp}{\frac{\partial f}{\partial x}+p\frac{\partial f}{\partial z}}=\frac{dq}{\frac{\partial f}{\partial y}+q\frac{\partial f}{\partial z}}$

Charpit's auxiliary equations for the given PDE:

$\frac{-dx}{q}= \frac{-dy}{p}= \frac{-dz}{2pq}=\frac{dp}{-p}=\frac{dq}{-q}$

One of the possible solutions is that with pq=0. In this case z=pq=0, $p=z_x=0$ and $q=z_y=0$ .

We assume from now that $pq\ne0$ .

$\frac{-dx}{q}=\frac{dq}{-q}$ implies that $dq=dx$, $q=x+a$

$\frac{-dy}{p}= \frac{dp}{-p}$ implies that $dp=dy$, $p=y+b$

Finally, z=pq=(x+a)(y+b).

The integral surface passing through C : x₀ = 0, y₀ = s, z₀ = s² must satisfy:

$s^2=a(b+s)$. Therefore, $a\ne 0$ and $b=a^{-1}s^2-s$.

Answer. $z=(x+a)(y+a^{-1}s^2-s)$