Question

p²+2pycotx=y²(p=dy/dx)
​

Answer 1

Answer:

Here's your answer..........

Step-by-step explanation:

2 + 2py cotx = y2.

(y')^2+2y'ycotx-y^2=0(y

′

)

2

+2y

′

ycotx−y

2

=0

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

′

)

2

+2y

′

ycotx−y

2

)/y

2

=0

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

′

/y)

2

+2y

′

cotx/y−1=0

z=y'/yz=y

′

/y

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

4/(sinx)

2

)/2=(1−cosx)/sinx

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

′

/y=(1−cosx)/sinx

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

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

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

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

2

=

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

2

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

4/(sinx)

2

)/2=−(1+cosx)/sinx

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

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

y(x)=c/(1-cosx)y(x)=c/(1−cosx)