# Answer to Question #223862 in Chemical Engineering for Lokika

Question #223862

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

1
2021-08-23T04:51:37-0400

p=zx,q=zy,

Let f(x,y,z,p,q)=yp2−2(z+xp+yq)

So that

fx=−2p,fy=p2−2q,fz=−2,fp=2py−2x,fq=−2y

Then we use Charpits method:

dx/fp=dy/fq=dz/(p*fp+q*fq)=−dp/(fx+p*fz)=−dq/(fy+q*fz)

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

we have p=c/(y*2) and −dp/(fx+p*fz)=−dq/(fy+q*fz)

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+(2yz)*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.

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