Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

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

share|improve this question
1  
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 –  Chico_Terry Jun 16 at 2:47
    
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). –  M. Vinay Jun 16 at 3:16
    
Similarly, sometimes you're given equations that can be reduced to separable form after an appropriate substitution. –  M. Vinay Jun 16 at 3:18

2 Answers 2

up vote 6 down vote accepted

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,

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

share|improve this answer
1  
Small correction: having $\ln x$ in the equation already tells us that $x \neq 0$. Instead, you should assume that $x \neq 1$ so $\ln x \neq 0$. –  Javier Badia Jun 16 at 2:50
    
Ah. This makes sense. Thanks for putting it really simply. :) –  Jesse Samson Jun 16 at 2:50
    
@JavierBadia, yeah I didn't realize that, I'll fix. –  Nameless Jun 16 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 $c\in \mathbb{R}.$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.