I have an exercise and I have an idea of how to solve it, could you help me by giving me a hint?
$$(10-6y+e^{-3x})d x- 2 d y=0, \qquad y(0)=1$$
This is my attemp
$$ \frac{\partial F}{\partial x} = P(x,y)\\ \frac{\partial F}{\partial y} = Q(x,y)$$ Is not exact, so find $\mu(x)$ $$ \frac{\frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x}}{Q} = 3 $$ Therefore $$ \mu (x) = e^{\int 3 dx} \implies e^{3x}$$ Solving $$ (10-6y+e^{-3x})dx- 2 dy=0\\ e^{3x}[ (10-6y+e^{-3x})dx- 2 dy=0]\\ (10e^{3x}-6ye^{3x}+1)dx-2e^{3x}dy = 0\\ 10\int e^{3x}dx-6y\int e^{3x}dx+\int dx = 2e^{3x} \int ydy$$ And in the last line is my doubt, I dont know how to separate the equation.
Thank you
Sometimes those problems that are written as though they could be exact differentials are actually trick questions, and it's easier to solve by a different method.
So in your case, I would divide through by $dx$ and rewrite the equation as $$y' +3y =\frac{ e^{-3x}}{2} +5.$$ Is this enough of a hint for you to know what to do to solve?