Answer to Question #262771 in Differential Equations for Torjan

Question

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

y²+z²=2,x+z=1

Answer 1

Solution;

Given

$4yzp+q+2y=0$ .....(1)

And;

$y^2+z^2=2$ and $x+z=1$ .....(2)

The Lagrange's auxiliary equations for (1) are:

$\frac{dx}{4yz}=\frac{dy}{1}=\frac{dz}{-2y}$ ......(3)

Taking the first and third equations in (3);

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

We have;

$dx+2zdz=0$

Integrate;

$x+z^2=c_1$ ....(4)

Now take the second and third fractions in (3) ;

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

We resolve to;

$dz+2ydy=0$

Integrate;

$z+y^2=c_2$ .....(5)

Add (4) and (5);

$(y^2+z^2)+(x+z)= c_1+c_2$ .....(6)

By substitution of (2);

$c_1+c_2=2+1=3$

Hence putting the value in (6),the required integral surface is;

$y^2+z^2+z+x-3=0$