## Calculus (3rd Edition)

$$y= \cos x \sin x -2\cos x .$$
This is a linear equation and has the integrating factor as follows $$\alpha(x)= e^{\int P(x)dx}=e^{ \int \tan x dx}=e^{-\ln \cos x}=\frac{1}{\cos x}$$ Now the general solution is \begin{align} y& =\alpha^{-1}(x)\left( \int\alpha(x) Q(x)dx +C\right)\\ & =\cos x \left( \int \cos x dx+C\right)\\ & =\cos x \left( \sin x+C\right)\\ & =\cos x \sin x +C\cos x \end{align} Since, $y(\pi)=2$, then $C=-2$ Thus, the general solution is: $$y= \cos x \sin x -2\cos x .$$