Question #294781

Solve (y ^ 2 * z)/x * p + xzq = y ^ 2 .

1
Expert's answer
2022-02-08T14:30:37-0500

y2zpy^2zp +x2zq=y2x+x^2zq=y^2x


The subsidiary equations are

dxy2z=dyx2z=dzy2x\frac{dx}{y^2z}=\frac{dy}{x^2z}=\frac{dz}{y^2x}


Considering the first two terms of the equation we get

dxy2=dyx2\frac{dx}{y^2}=\frac{dy}{x^2}


x2dxy2dy=0x^2dx-y^2dy=0

x3y3=c1x^3-y^3=c_1


Considering the 1st and last terms of the equation

dxy2z=dzy2x\frac{dx}{y^2z}=\frac{dz}{y^2x}


dxz=dzx\frac{dx}{z}=\frac{dz}{x}


xdxzdz=0xdx-zdz=0

x2z2=c2x^2-z^2=c_2


Solution


f(x3y3,x2z2)=0f(x^3-y^3,x^2-z^2)=0


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