Question

Find the complete integral of partial differential equations z+xp-x²yq²-x³pq =0

Answer 1

f(x,y,z,p,q)=xpq+yq2-1=0$\frac{dp}{\frac{df}{dx}+p\frac{df}{dz}}=\frac{dq}{\frac{df}{dy}+q\frac{df}{dz}}=\frac{dz}{-p\frac{df}{dp}-q\frac{df}{dq}}=\frac{dx}{-\frac{df}{dp}}=\frac{dy}{-\frac{df}{dq}}$

$\frac{dp}{p-2xyq^2-3x^2pq+p}=\frac{dq}{-x^2q^2+q}=\frac{dz}{-p(x-x^3q)-q(2qyx^2-x^3p)}=\frac{dx}{-x+x^3q}=\frac{dy}{2qyx^2+x^3p}$

$\frac{\frac{dq}{q}}{1-qx^2}=\frac{\frac{dx}{x}}{x^2q-1}$

$\frac{dq}{q}+ \frac{dx}{x}=0$

ln(q)+ln(x)=a

qx=e^a

q=e^a/x

$\frac{ \frac{dp}{qp}+ \frac{dy}{xp}}{\frac{2}{q} - \frac{2xyq}{p}-3x^2+ \frac{2qyx}{p}+ x^2}=\frac{ \frac{dp}{qp}+ \frac{dy}{xp}}{\frac{2}{q} -2x^2}=0$

$\frac{dp}{q}+ \frac{dy}{x}=0$

$\frac{p}{q}=- \frac{y}{x} + b$

$\frac{e^ap}{x}=- \frac{y}{x} + b$

p=-yb/e^a

dz=pdx+qdy=-yb/e^adx+e^a/xdy

z=-ybxe^a+e^ay/x+c=e^ayx (1/x² - b)+c