Answer to Question #128667 in Differential Equations for Duaa

Question #128667
Find the complete integral of the following equation by
the Charpits Method 2(z + px + yq) = yp^2
.
1
Expert's answer
2020-08-12T18:00:53-0400

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/(pfp+qfq)=−dp/(fx+pfz)=−dq/(fy+qfz)

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

we have p=c/(y*2) and from −dp/(fx+pfz)=−dq/(fy+qfz)

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

Finally get next:

dz=(c/y2)dx+((c2−2y3z−2cxy2)/(2y4))*dy.

we have the equation :

dz=(c/y2)dx+((c2−2y3z−2cxy2)/(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+ c2dy/(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+2c(ydx−xdy)/(y2)

⟹2(ydz+zdy)=(c2/y3)dy+2c(ydx−xdy)/(y2)

(ydz+zdy) is a differential of yz (differential of product)

(ydx−xdy)/(y2) is a differential of x/y

So we have: 2d(zy)=(c2/y3)dy+2cd(x/y)

⟹2d(zy)=(c2/y3)dy+2cd(x/y)

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

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


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS