Substitution in the DE $y' \tan(y)+4x^3\cos(y)=2x$

Question

I have to solve the differential equation, $$ y' \tan(y)+4x^3\cos(y)=2x$$

I propose the substitution $u=\cos(y)$, so $$u'=-y'\sin(y) \\\Rightarrow \sin(y) = -\dfrac{u'}{y'} \\\Rightarrow \tan(y)=\dfrac{\sin(y)}{\cos(y)}=-\dfrac{u'}{y'}\dfrac{1}{u}$$

Then substituting into the DE, gives

$$-y' \dfrac{u'}{y'}\dfrac{1}{u}+4x^3u=2x \\ \Rightarrow -\dfrac{u'}{u}+4x^3u=2x \\ \Rightarrow u'-4x^3u^2+2xu=0$$

Which clearly, is a Bernoulli equation.

My question is: Is it correct to use the substitution that way?

I have doubts because I have never used a substitution like that

Answer 1

Possible a simplification: divide by $\cos(y)$. You'll have that:

$$y' \sec(y) \tan(y) -2x\sec(y) = -4x^3$$ $$\dfrac{d}{dx}\sec(y) -2x\sec(y) = -4x^3$$

Now its directly seen to be a linear differential equation.

And for your question about substitution, its fine. It just means you are calling the function by some other name. If you prefer, you can just leave it in original form.