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.

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

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    $\begingroup$ maybe if you divide by xln(x) on both sides you can use integrating factor. You can also check your final answer here: onsolver.com/diff-equation.php $\endgroup$ – Chico_Terry Jun 16 '14 at 2:47
  • $\begingroup$ The standard method is to rearrange it into a linear equation and solve (as done in the answers). What your professor wanted you to do, probably, is rearrange it slightly by (perhaps) noticing the $x$ multiplied with two terms and dividing by it to get $\ln(x) \dfrac{dy}{dx} + \dfrac{y}{x} = e^x$, and now observing that this can be written as $\dfrac{d(y\ln(x))}{dx} = e^x$, which is in "separable" form, so $y\ln(x) = \int e^x dx + C$. This is exactly what the linear equation solution does. This problem might have been used as a way to foreshadow linear equations. (I do that often too). $\endgroup$ – M. Vinay Jun 16 '14 at 3:16
  • $\begingroup$ Similarly, sometimes you're given equations that can be reduced to separable form after an appropriate substitution. $\endgroup$ – M. Vinay Jun 16 '14 at 3:18

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,

$$y = \frac{e^x}{\ln x} + \frac{C}{\ln x}. $$

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    $\begingroup$ Small correction: having $\ln x$ in the equation already tells us that $x \neq 0$. Instead, you should assume that $x \neq 1$ so $\ln x \neq 0$. $\endgroup$ – Javier Jun 16 '14 at 2:50
  • $\begingroup$ Ah. This makes sense. Thanks for putting it really simply. :) $\endgroup$ – Jesse Samson Jun 16 '14 at 2:50
  • $\begingroup$ @JavierBadia, yeah I didn't realize that, I'll fix. $\endgroup$ – IAmNoOne Jun 16 '14 at 2:51

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 $c\in \mathbb{R}.$


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