Answer to Question #178192 in Differential Equations for m siva prasad

Question

Solve yz² (x²-yz)dx+zx² (y²-zx)dy+xy² (z²-xy)dz=0

Answer 1

The condition of integrability:

$P(\frac{\partial Q}{\partial z}-\frac{\partial R}{\partial y})+Q(\frac{\partial R}{\partial x}-\frac{\partial P}{\partial z})+R(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x})=0$

$(yx^2z^2-y^2z^3)(x^2y^2-2zx^3-2yxz^2+3y^2x^2)+(zx^2y^2-z^2x^3)(y^2z^2-2xy^3-$

$-2zyx^2+3z^2y^2)+(xy^2z^2-x^2y^3)(x^2z^2-2yz^3-2xzy^2+3x^2z^2)=$

$=y^3x^4z^2-2yx^5z^3-2y^2x^3z^4+3y^3x^4z^2-x^2y^4z^3+2x^3y^2z^4+2xy^3z^5-3x^2y^4z^3+$

$+x^2y^4z^3-2x^3y^5z-2x^4y^3z^2+3x^2y^4z^3-x^3y^2z^4+2x^4y^3z^2+2x^5yz^3-3x^3y^2z^4+$

$+x^3y^2z^4-2xy^3z^5-2x^2y^4z^3+3x^3y^2z^4-x^4y^3z^2+2x^2y^4z^3+2x^3y^5z-3x^4y^3z^2=$

$=0$

The equation is integrable.

The equation is homogeneous.

$D=Px+Qy+Rz=xyz^2(x^2-yz)+yzx^2(y^2-zx)+zxy^2(z^2-xy)=$

$=x^3yz^2-xy^2z^3+x^2y^3z-x^3yz^2+xy^2z^3-x^2y^3z=0$

Let

$x=uz,y=vz$

$dx=udz+zdu,dy=vdz+zdv$

Then:

$vz^3(u^2z^2-vz^2)(udz+zdu)+z^3u^2 (v^2z^2-z^2u)(vdz+zdv)+$

$+uv^2z^3(z^2-uvz^2)dz=0$

$(vu^3-v^2u)dz+(vu^2z-v^2z)du+(v^3u^2-vu^3)dz+(zv^2u^2-zu^3)dv+$

$+(uv^2-u^2v^3)dz=0$

$vz(u^2-v)du+u^2z(v^2-u)dv=0$

$v(u^2-v)du+u^2(v^2-u)dv=0$

$\frac{dv}{du}=\frac{v(u^2-v)}{u^2(u-v^2)}$

$v=\frac{\sqrt{(c+1)^2-4u^3}+cu+1}{2u}$

$\frac{y}{z}=\frac{z(\sqrt{(c+1)^2-4(x/z)^3}+(cx/z)+1)}{2x}$