Find the general solution to the differential equation. Primes denote derivatives with respect to x throughout. (2xsin(y)cos(y))y' = 4x^2 + (sin(y))^2
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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.
I was able to get this far but I am still having trouble. Should the integrating factor be -(1/x)
Originally Posted by bearej50 I was able to get this far but I am still having trouble. Should the integrating factor be -(1/x) It's close. You have a sign error. $\displaystyle u'x=4x^2+u$ $\displaystyle u'x - u =4x^2$ $\displaystyle u' - \frac{u}{x} =4x$ $\displaystyle IF = e^{-\int \frac{1}{x} \ \mathrm{d}x} = e^{-\ln x} = e^{\ln |x^{-1}|} = x^{-1} = \frac{1}{x}$
Any multiple of an integrating factor suffices. That's why you don't need the +C when you integrate.
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