Question

𝑥𝑝𝑞 + 𝑦q² = 1

Answer 1

f(x,y,z,p,q)=xpq+yq²-1=0

$\frac{dp}{\frac{df}{dx}+p\frac{df}{dz}}=\frac{dq}{\frac{df}{dy}+q\frac{df}{dz}}=\frac{dz}{-p\frac{df}{dp}-q\frac{df}{dq}}=\frac{dx}{-\frac{df}{dp}}=\frac{dy}{-\frac{df}{dq}}$

$\frac{dp}{pq}=\frac{dq}{q^2}=\frac{dz}{-pxq-2yq^2-qxp}=\frac{dx}{-xq}=\frac{dy}{-xp-2yq}$

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

$ln(q)+ln(x)=a$

ln(qx)=a

q=e^a/x

$\frac{dp}{p}- \frac{dq}{q}=0$

ln(p)-ln(q)=b

ln(p/q)=b

p=e^bq=e^(a+b)/x

dz=pdx+qdy

dz=(e^(a+b)/x)dx+(e^a/x)dy

z=e^(a+b)ln(x)+(e^ay/x)+c