Obtain the complete integral xp^2- ypq + y^3q - y^2z= 0
Using charpit's auxilliary equation,
"\\frac{dp}{\u2202f\/\u2202x + p(\u2202f\/\u2202z)}=\\frac{dq}{\u2202f\/\u2202y + q(\u2202f\/\u2202z)}=\\frac{dz}{\u2212p(\u2202f\/\u2202p)\u2212 q(\u2202f\/\u2202q)}=\\frac{dx}{\u2212\u2202f\/\u2202p}=\\frac{dy}{\u2212\u2202f\/\u2202q}\\\\\n\n\\frac{\u2202f}{\u2202x}=p^2\\\\\n\\frac{\u2202f}{\u2202z}=-y^{2}\\\\\n\\frac{\u2202f}{\u2202y}=-pq+3qy^2-2zy\\\\\n\\frac{\u2202f}{\u2202p}=2xp-yq\\\\\n\\frac{\u2202f}{\u2202q}=-yp+y^3\\\\\n\n\\text{so by substitution}\\\\\n\n\\frac{dp}{p^2-py^2}=\\frac{dq}{2qy^2-pq-2zy}=\\frac{dz}{-2xp^2+2ypq-qy^3}=\\frac{dx}{yq-2xp}=\\frac{dy}{yp-y^3}\\\\\n\n\\frac{dp}{p}-\\frac{dy}{y}=0\\\\\nlog p-logy=log a\\\\\nlog p=log a+logy\\\\\np=ay\\\\\n\\text{substitute the value of p in the equation and we have q to be}\\\\\n\nx({ay})^2-y(ay)q+q({ay})^3-zy^2=0\\\\\nq=\\frac{z-xa^2}{ay-a}\\\\\n\\text{So we use}\\\\ \ndz=pdx+qdy\\\\\ndz=aydx+ \\frac{z-xa^2}{ay-a}dy\\\\\n\\text{by integrating both sides, we have}\\\\\nz=axy+(z-xa^2)log_e(ay-a)\\\\\nz=axy+blog_e(ay-a)\\\\\n\\text{Where a and b are arbitrary constants}"
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