Answer to Question #128667 in Differential Equations for Duaa

Question

Find the complete integral of the following equation by
the Charpits Method 2(z + px + yq) = yp^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/(pf_p+qf_q)=−dp/(f_x+pf_z)=−dq/(f_y+qf_z)

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

we have p=c/(y*2) and from −dp/(f_x+pf_z)=−dq/(f_y+qf_z)

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+(2y³ z)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.