Find the general solution to the differential equation. Primes denote derivatives with respect toxthroughout.

(2xsin(y)cos(y))y' = 4x^2 + (sin(y))^2

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- Feb 20th 2009, 06:57 AMbearej50Bernoulli Differential Equation
Find the general solution to the differential equation. Primes denote derivatives with respect to

*x*throughout.

(2*x*sin(*y*)cos(*y*))*y*' = 4*x*^2 + (sin(*y*))^2

- Feb 20th 2009, 07:20 AMKrizalid
Make the substitution $\displaystyle u=\sin^2y$ so that $\displaystyle u'=2\sin(y)\cos(y)y'$ and the ODE is $\displaystyle u'x=4x^2+u$ which is linear.

- Feb 20th 2009, 09:35 AMbearej50
I was able to get this far but I am still having trouble. Should the integrating factor be -(1/

*x*) - Feb 20th 2009, 11:02 AMAir
- Feb 22nd 2009, 09:25 PMmatheagle
Any multiple of an integrating factor suffices.

That's why you don't need the +C when you integrate.