Answer to Question #151075 in Differential Equations for Ashweta Padhan

Question

Find a integral surface of the differential equation p^2x+ pqy =2pz+x, passing through the line y=1, x=z

Answer 1

p²x+pqy-2pz-x=0

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

$\frac{dp}{p^2-1+p+-2p)}=\frac{dq}{pq+q(-2p)}=\frac{dz}{-p(2px+qy-2z)-qpy}=\frac{dx}{-(2px+qy-2z)}=\frac{dy}{-py}$

$\frac{dp}{-p^2-1}=\frac{dq}{-pq}=\frac{dz}{-2p^2x-2pqy+2pz}=\frac{dx}{-2px-qy+2z}=\frac{dy}{-py}$

$\frac{dp}{-p^2-1}=0$

-arctan(p)=a

p=tan(-a)=c

-pqdp=(-p²-1)dq

$\frac{-p}{-p^2-1}dp=\frac{dq}{q}$

Let t=p², dt=2p

$\int{\frac{-p}{-p^2-1}dp}=-\int{\frac{1}{2(-t-1)}dt}=\frac{1}{2}\int{\frac{1}{t+1}dp}=\frac{ln(t+1)}{2}=\frac{ln(p^2+1)}{2}$

$\frac{ln(p^2+1)}{2}=lnq$

p²+1=q²

$q=\sqrt{p^2+1}=\sqrt{c^2+1}$

dz=pdx+qdy

$dz=cdx+\sqrt{c^2+1}dy$

$z=cx+\sqrt{c^2+1}y+b$

y=1, x=z

$z=cz+\sqrt{c^2+1}+b$

$z(1-c)=\sqrt{c^2+1}+b$

$z=\frac{\sqrt{c^2+1}+b}{1-c}$

Answer:$c^2x+c\sqrt{c^2+1}y=2cz+x$