Question #88757
Solve p2 + 2py cotx = y2.
1
Expert's answer
2019-04-29T10:20:40-0400
(y)2+2yycotxy2=0(y')^2+2y'ycotx-y^2=0

((y)2+2yycotxy2)/y2=0((y')^2+2y'ycotx-y^2)/y^2=0

(y/y)2+2ycotx/y1=0(y'/y)^2+2y'cotx/y-1=0

z=y/yz=y'/y

z2+2zcotx1=0z^2+2zcotx-1=0

D=(2cotx)2+4=4(1+(cotx)2)=4/(sinx)2D=(2cotx)^2+4=4(1+(cotx)^2)=4/(sinx)^2

So we have two cases.


1)


z=(2cotx+4/(sinx)2)/2=(1cosx)/sinxz=(-2cotx+\sqrt{4/(sinx)^2})/2=(1-cosx)/sinx

y/y=(1cosx)/sinxy'/y=(1-cosx)/sinx

dy/y=(1cosx)dx/sinxdy/y=(1-cosx)dx/sinx

dy/y=(1cosx)dx/sinx\intop dy/y=\intop (1-cosx)dx/sinx

lny=(1cosx)dx/sinxlny=\intop (1-cosx)dx/sinx

(1cosx)dx/sinx=(1cosx)d(cosx)/(sinx)2=\intop (1-cosx)dx/sinx=-\intop (1-cosx)d(cosx)/(sinx)^2=

=(1cosx)d(cosx)/(1(cosx)2)=d(cosx)/(1+cosx)==-\intop (1-cosx)d(cosx)/(1-(cosx)^2)=-\intop d(cosx)/(1+cosx)=

=ln(1+cosx)+C=-ln(1+cosx)+C

So we get:



lny=ln(1+cosx)+C=ln(1/(1+cosx))+Clny=-ln(1+cosx)+C=ln(1/(1+cosx))+C

y(x)=c/(1+cosx)y(x)=c/(1+cosx)

2)


z=(2cotx4/(sinx)2)/2=(1+cosx)/sinxz=(-2cotx-\sqrt{4/(sinx)^2})/2=-(1+cosx)/sinx

lny=(1+cosx)dx/sinx=ln(1cosx)+Clny=-\intop (1+cosx)dx/sinx=-ln(1-cosx)+C

lny=ln(1/(1cosx)+Clny=ln(1/(1-cosx)+C

y(x)=c/(1cosx)y(x)=c/(1-cosx)


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