How to solve the following equation? $$y' + y^2 - 2y\sin x + \sin^2x = \cos x$$

It is necessary to determine the type and total solution. Help me please.


$$y' + y^2 - 2y\sin x + \sin^2x = \cos x$$ $$(y' - \cos x) + (y-\sin x)^2 = 0$$

Say $z=y- \sin x$
Then $\frac{dz}{dx}= y'-\cos x$

So we have $$\frac{dz}{dx}+z^2=0$$ $$\int z^{-2} dz = -\int dx$$ $$\frac{1}{z}=x+c$$ $$\frac{1}{y- \sin x}=x+c$$ $$y = \sin x + \frac{1}{x+c}$$

This is the required solution.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.