Question

solve the Bernoulli equation dy/dx+ytanx=y^3secx​

Answer 1

Step-by-step explanation:

Solve

dydx+ytanx=y3secx

My Attempt:

dydx+ytanx=y3secx

Dividing both sides y3,

y−3dydx+y−2tanx=secx

Put y−2=z

(−2)y−2dydx=dzdx

y−3dydx=−12dzdx

Then

−12dzdx+z.tanx=secx

−dzdx+(2tanx)z=2secx

y′y3+tanxy2=1cosx

z′−2ztanx=−2cosx

The integrating factor would be now:

e−2∫tanxdx=eln(cos2x)+C

So multiply both sides by cos2x

(zcos2x)′=−2cosx⇒zcos2x=−2sinx+C

y=cos2x−2sinx+C−−−−−−−−−−−√