Question

px(z-2y^2)=(z-qy)(z-y^2-2x^3)​

Answer 1

Answer:

p=∂z∂x and q=∂z∂y

Find the general integral of the linear PDE px(z−2y2)=(z−qy)(z−y2−2x3).

My attempt to solve this is as follows: p=∂z∂x and q=∂z∂y

px(z−2y2)+qy(z−y2−2x3)=z(z−y2−2x3)

The Lagrange's auxiliary equation is:dxx(z−2y2)=dyy(z−y2−2x3)=dzz(z−y2−2x3)

Now consider the 2nd and 3rd ratios,

dyy(z−y2−2x3)⟹dyy⟹ln(y)⟹yz=dzz(z−y2−2x3)=dzz=ln(z)+ln(c1)=c1.

Answer 2

Answer-

Step-by-step explanation:

given px(z-2y^2)=(z-qy)(z-y^2-2x^3) can be written as

px(z-2y^2)+qy(z-y^2x^3)=z(z-y^2-2x^3)

Auxiliary equation of above equation is ;

dx/P=dy/Q=dz/R ;

where P=x(z-2y^2),Q=y(z-y^2x^3),R=z(z-y^2-2x^3) ;

from dy/Q=dz/R ;

dy/y=dz/z ;

by apply integration on both sides to above equation we get [y/z]=c1 ;

from dx/P=(dz-2ydy)/(R-Q) ;

dx/x =(dz-2ydy)/(z-y^2-2x^3) ; ______eq 1 ;

let z-y^2=t;

dz-2ydy=dt ______eq 2 ;

substitute eq 2 in eq 1 ;

(t-2x^3)dx=xdt ;

tdx-xdt=2x^3dx ;

-[(xdt-tdx)/x^2]=2xdx ;

-d[t/x]=2xdx _____ eq 3 ;

by applying integration to above eq 3 we get

x^2+(z-y^2/x) = c2;

the required solution is f(y/z,x^2+(z-y^2/x))=0