p2 x+pqy-2pz-x=0
d p d f d x + p d f d z = d q d f d y + q d f d z = d z − p d f d p − q d f d q = d x − d f d p = d y − d f d q \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}} d x df + p d z df d p = d y df + q d z df d q = − p d p df − q d q df d z = − d p df d x = − d q df d y
d p p 2 − 1 + p + − 2 p ) = d q p q + q ( − 2 p ) = d z − p ( 2 p x + q y − 2 z ) − q p y = d x − ( 2 p x + q y − 2 z ) = d y − p y \frac{dp}{p^2-1+p+-2p)}=\frac{dq}{pq+q(-2p)}=\frac{dz}{-p(2px+qy-2z)-qpy}=\frac{dx}{-(2px+qy-2z)}=\frac{dy}{-py} p 2 − 1 + p +− 2 p ) d p = pq + q ( − 2 p ) d q = − p ( 2 p x + q y − 2 z ) − qp y d z = − ( 2 p x + q y − 2 z ) d x = − p y d y
d p − p 2 − 1 = d q − p q = d z − 2 p 2 x − 2 p q y + 2 p z = d x − 2 p x − q y + 2 z = d y − p y \frac{dp}{-p^2-1}=\frac{dq}{-pq}=\frac{dz}{-2p^2x-2pqy+2pz}=\frac{dx}{-2px-qy+2z}=\frac{dy}{-py} − p 2 − 1 d p = − pq d q = − 2 p 2 x − 2 pq y + 2 p z d z = − 2 p x − q y + 2 z d x = − p y d y
d p − p 2 − 1 = 0 \frac{dp}{-p^2-1}=0 − p 2 − 1 d p = 0
-arctan(p)=a
p=tan(-a)=c
-pqdp=(-p2 -1)dq
− p − p 2 − 1 d p = d q q \frac{-p}{-p^2-1}dp=\frac{dq}{q} − p 2 − 1 − p d p = q d q
Let t=p2 , dt=2p
∫ − p − p 2 − 1 d p = − ∫ 1 2 ( − t − 1 ) d t = 1 2 ∫ 1 t + 1 d p = l n ( t + 1 ) 2 = l n ( p 2 + 1 ) 2 \int{\frac{-p}{-p^2-1}dp}=-\int{\frac{1}{2(-t-1)}dt}=\frac{1}{2}\int{\frac{1}{t+1}dp}=\frac{ln(t+1)}{2}=\frac{ln(p^2+1)}{2} ∫ − p 2 − 1 − p d p = − ∫ 2 ( − t − 1 ) 1 d t = 2 1 ∫ t + 1 1 d p = 2 l n ( t + 1 ) = 2 l n ( p 2 + 1 )
l n ( p 2 + 1 ) 2 = l n q \frac{ln(p^2+1)}{2}=lnq 2 l n ( p 2 + 1 ) = l n q
p2 +1=q2
q = p 2 + 1 = c 2 + 1 q=\sqrt{p^2+1}=\sqrt{c^2+1} q = p 2 + 1 = c 2 + 1
dz=pdx+qdy
d z = c d x + c 2 + 1 d y dz=cdx+\sqrt{c^2+1}dy d z = c d x + c 2 + 1 d y
z = c x + c 2 + 1 y + b z=cx+\sqrt{c^2+1}y+b z = c x + c 2 + 1 y + b
y=1, x=z
z = c z + c 2 + 1 + b z=cz+\sqrt{c^2+1}+b z = cz + c 2 + 1 + b
z ( 1 − c ) = c 2 + 1 + b z(1-c)=\sqrt{c^2+1}+b z ( 1 − c ) = c 2 + 1 + b
z = c 2 + 1 + b 1 − c z=\frac{\sqrt{c^2+1}+b}{1-c} z = 1 − c c 2 + 1 + b
Answer:c 2 x + c c 2 + 1 y = 2 c z + x c^2x+c\sqrt{c^2+1}y=2cz+x c 2 x + c c 2 + 1 y = 2 cz + x
Comments
Using that differential equation one can conclude p is constant. It can be proved because pdp/(p^2+1)=dq/q=dy/y. If the function of p simultaneously depends on different variables, it is constant.
How you have taken dp/(p^2)-1=0 give reason