Question #223862

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

Expert's answer

p=z_{x},q=z_{y},

Let f(x,y,z,p,q)=yp^{2}−2(z+xp+yq)

So that

f_{x}=−2p,f_{y}=p^{2}−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=p^{3}/(12+a), where a and c are constants.

Finally get the next expression:

dz=(c/y^{2})*dx+((c^{2}−2*y^{3}*z−2*c*x*y^{2})/2y^{4})*dy.

we have the equation :

dz=(c/y^{2})*dx+((c^{2}−2*y^{3}*z−2*c*x*y^{2})/2y^{4})*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+(2y^{3 }z)*dy/2y^{4}=(с/у^{2})*dx+ c^{2}*dy/2y^{4}- 2cxydy/2y^{4}

Then we get: dz+(zdy)/y=cdx/y^{2}+c^{2}dy/2y^{4}-cxdy/y^{3}

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

2(ydz+zdy)=(c^{2}/y^{3})*dy+2*c*(y*dx−x*dy)/(y^{2})

⟹2(ydz+zdy)=(c^{2}/y^{3})*dy+2*c*(y*dx−x*dy)/(y^{2})

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

(y*dx−x*dy)/(y^{2}) is a derivative of x/y

So we have: 2d(z*y)=(c^{2}/y^{3})*dy+2*c*d*(x/y)

⟹2d(z*y)=(c^{2}/y^{3})*dy+2*c*d*(x/y)

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

⟹2zy=−c^{2}/(2y^{2})+2cx/y+c_{1}, where c_{1} is constant.

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