Is there a special way to solve a problem like this with the indicated setup?

Question

My Answer, and the Question

My question is if the problem was converted to the following:\begin{align}\cos(x)y'+\sin(x)y&=2\sin(x)\\ u'y'+uy&=2u \end{align} How can I advance from here to solve the problem if substitution were my aim in answering the question (which of course is answered)?

Answer 1

Your approach, $$u'y'+uy=2u$$ $$uy'-u'y=-2u'\quad (\text{where, } \ \ u=\cos x)$$ $$\frac{y'}{u}-\frac{u'}{u^2}y=-2\frac{u'}{u^2}\quad (\text{dividing by } \ u^2)$$ $$d\left(\frac{y}{u}\right)=2 \ d\left(\frac{1}{u}\right)$$ $$\int d\left(\frac{y}{u}\right)=2\int d\left(\frac{1}{u}\right)$$ $$\frac yu=\frac2u+C$$ $$y=Cu+2$$

Alternatively, $$\cos(x)y'+\sin (x)y=2\sin(x)$$ $$y'+\tan (x)y=2\tan(x)$$ Multiplying $\sec x$ on both the sides, $$y'\sec (x)+\sec(x)\tan (x)y=2\sec(x)\tan(x)$$ $$\frac{d}{dx}\left(y\sec (x)\right)=2\sec(x)\tan(x)$$ $$\int d\left(y\sec (x)\right)=\int 2\sec(x)\tan(x)dx$$ $$y\sec(x)=2\sec(x)+C$$ $$y=C\cos (x)+2$$

Answer 2

$$\begin{align}\cos(x)y'+\sin(x)y&=2\sin(x)\\ u'y'+uy&=2u \end{align}$$ Then you can do nothing. It's better to write it this way: $$\begin{align}uy'-u'y&=2\sin(x) \end{align}$$ $$\begin{align}uy'-u'y&=-2u'\end{align}$$ Where $u=\cos x$. Divide by$u^2$ both sides and integrate. $$\dfrac yu=-2\int \dfrac {du}{ u^2}$$