I can't quite figure out this "separable equation" My prof assigned this question for exam studying and I can't figure it out. It's supposed to be a separable equations question and I'd be able to do something, but for that pesky '$+ y$'.
All we've learned so far is separable equations and I feel like this is something more.

$x\ln(x) \dfrac{dy}{dx} + y = xe^x$

 A: I don't think you are suppose to seperate this DE.
Instead, assuming $\ln x \neq 0$, divide $x\ln x$ throughout the equation and get
$$y' + \frac{1}{x \ln x}y = \frac{e^x}{\ln x}.$$
Multiply both sides by the integrating factor $p = e^{\int \frac{dx}{x \ln x}} = \ln(x)$ to get, $$(\ln x)y' + \frac{1}{x}y = e^x.$$
Or equivalently,  $$(y\ln x)' = e^x.$$
Integrating both sides and dividing $\ln x$ to get,
$$y = \frac{e^x}{\ln x} + \frac{C}{\ln x}. $$
A: This ODE cannot be solved using separation of variables. Instead, we need to put this ODE into the following form $${dy\over dx}+p(x)y=q(x).$$ Consider $${dy\over dx}+{1\over xln(x)}y={e^x\over ln(x)}.$$ We need to make use of the integrating factor $I(x)=e^{\int p(x)dx}=e^{{1\over xln(x)}dx}=ln(x)$.Multiplying both sides of our ODE by $I(x)$ we obtain $$ln(x){dy\over dx}+{1\over x}y=e^x.$$ We can rewrite the above ODE as $$(y\cdot ln(x))'=e^x.$$ Integrating both sides with respect to $x$ we obtain $$y={e^x\over ln(x)}+{C\over ln(x)}$$ where $c\in \mathbb{R}.$
