Let's rewrite the equation in form
lny=xyp+y2p2
Denote Y=lny hence P=dxdY=yp
Terefore
Y=xP+P2
which is in the Clairaut's form
The solution can be obtained by just replacing P by constant c.
Y=cx+c2
or
lny=cx+c2
y=exp(cx+c2)
To find a singular solution, let's solve the equation
(P2)′+x=0
2P=−x
P=−2x
Therefore
Y=−2x2+(−2x)2=−41x2
will be the singular solution of this equation
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