Question

Find the integral surface of the equation 4yzp+q+2y=0 passing through

y²+z²=2,x+z=1

Answer 1

Characteristic system

$\frac{dx}{4yz}=\frac{dy}{1}=\frac{dz}{-2y}$

First and last terms can be rewritted as

$\frac{dx}{4yz}=\frac{-2zdz}{4yz}$

Therefore

dx=--2zdz

$d(x+z^2)=0; x+z^2=C_1-first\space integral$

Second and third terms of the characteristic system give:

-2ydy=dz;

$d(z+y^2)=0; z+y^2=C_2-second\space integral$

General solution may be written in the form^

$x+z^2=F(z+y^2)$

where F(t)-some function

Let $y^2+z^2=2,z+x=1-$ initial curve

Then $z+y^2=2+z-z^2,x+z^2=1-z+z^2$

Must be:

$1-z+z^2=F(2+z-z^2)=F(3-(1-z+z^2))\space \forall z$

or

F(3-t)=t=-(3-t)+3? therefore F(u)=3-u

Therefore surface equation is:

$x+z^2=3-z-y^2;\\ Answer:\\ z^2+y^2+z+x=3$