Solve p2 + 2py cotx = y2.
1
2019-04-29T10:20:40-0400
(y′)2+2y′ycotx−y2=0
((y′)2+2y′ycotx−y2)/y2=0
(y′/y)2+2y′cotx/y−1=0
z=y′/y
z2+2zcotx−1=0
D=(2cotx)2+4=4(1+(cotx)2)=4/(sinx)2 So we have two cases.
1)
z=(−2cotx+4/(sinx)2)/2=(1−cosx)/sinx
y′/y=(1−cosx)/sinx
dy/y=(1−cosx)dx/sinx
∫dy/y=∫(1−cosx)dx/sinx
lny=∫(1−cosx)dx/sinx
∫(1−cosx)dx/sinx=−∫(1−cosx)d(cosx)/(sinx)2=
=−∫(1−cosx)d(cosx)/(1−(cosx)2)=−∫d(cosx)/(1+cosx)=
=−ln(1+cosx)+C So we get:
lny=−ln(1+cosx)+C=ln(1/(1+cosx))+C
y(x)=c/(1+cosx)
2)
z=(−2cotx−4/(sinx)2)/2=−(1+cosx)/sinx
lny=−∫(1+cosx)dx/sinx=−ln(1−cosx)+C
lny=ln(1/(1−cosx)+C
y(x)=c/(1−cosx)
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