Answer to Question #223862 in Chemical Engineering for Lokika

Question

Find the complete integral of xp +3yq = 2(z-x^2.q^2).

Answer 1

p=z_x,q=z_y,

Let f(x,y,z,p,q)=yp²−2(z+xp+yq)

So that

f_x=−2p,f_y=p²−2q,f_z=−2,f_p=2py−2x,f_q=−2y

Then we use Charpits method:

dx/f_p=dy/f_q=dz/(p*f_p+q*f_q)=−dp/(f_x+p*f_z)=−dq/(f_y+q*f_z)

Then putting all the values and then equating second and fourth terms

we have p=c/(y*2) and −dp/(f_x+p*f_z)=−dq/(f_y+q*f_z)

gives p*q=p³/(12+a), where a and c are constants.

Finally get the next expression:

dz=(c/y²)*dx+((c²−2*y³*z−2*c*x*y²)/2y⁴)*dy.

we have the equation :

dz=(c/y²)*dx+((c²−2*y³*z−2*c*x*y²)/2y⁴)*dy

let's open the brackets on the right side of the equation and then

move to the left part of the expression written to the right, so we have:

dz+(2y³ z)*dy/2y⁴=(с/у²)*dx+ c²*dy/2y⁴- 2cxydy/2y⁴

Then we get: dz+(zdy)/y=cdx/y²+c²dy/2y⁴-cxdy/y³

Now we multiply all expressions by 2y, after that we get

2(ydz+zdy)=(c²/y³)*dy+2*c*(y*dx−x*dy)/(y²)

⟹2(ydz+zdy)=(c²/y³)*dy+2*c*(y*dx−x*dy)/(y²)

(ydz+zdy) is a derivative of y*z (derivative of product)

(y*dx−x*dy)/(y²) is a derivative of x/y

So we have: 2d(z*y)=(c²/y³)*dy+2*c*d*(x/y)

⟹2d(z*y)=(c²/y³)*dy+2*c*d*(x/y)

and then we integrate this equation and in result we have:

⟹2zy=−c²/(2y²)+2cx/y+c₁, where c₁ is constant.