## anonymous 5 years ago Can someone help me solve this Differential equation? dy+lnxy dx=(4x+lny)dx

1. anonymous

I think this is where I start? $dy/dx+\ln xy=4x+\ln y$

2. anonymous

You just have to use log laws to open the product of xy in the LHS log into a sum. That is,$\frac{dy}{dx}+\ln xy = 4x + \ln y \rightarrow \frac{dy}{dx}+\ln x + \ln y = 4x + \ln y$The logs of y cancel and you're left with$\frac{dy}{dx}=4x-\ln x \rightarrow y=2x^2-x \ln x +x+c$

3. anonymous

$=2x^2+x(1-\ln x)+c$

4. anonymous

Do you know how to integrate ln(x)?

5. anonymous

Yes, I'm not too familiar with the laws of logs thats where I needed help. Does dy just = y?

6. anonymous

Oh, are you asking if dy becomes y as in $\frac{dy}{dx}=4x- \ln x \rightarrow \int\limits_{}{}dy=\int\limits_{}{}4x-\ln x dx \rightarrow$$y=2x^2-x \ln x + x+c$?

7. anonymous

If so, yes.

8. anonymous

Right... and isn't $\int\limits_{}^{} lnx dx = x-\ln(x)-x$ ? You have +x

9. anonymous

No...$\int\limits_{}{}\ln x dx=x \ln x -x+c$I have a + because I skipped a step: when you sub. the result of the integral in, you'll have$2x^2-(x \ln x -x) +c$which is equal to $2x^2-x \ln x + x +c$

10. anonymous

You can integrate ln(x) using integration by parts, taking u=ln(x), dv=dx and going from there.

11. anonymous

Ohh I see I forgot to distribute the - back into it.

12. anonymous

Yep

13. anonymous

You can always check your solution by subbing it into the differential equation.

14. anonymous

Ok and for the final anwer you just moved the x on the back, to the front and factored a little right?

15. anonymous

Yeah, for the final answer (which technically you don't need to do since the first answers the question), I've just taken out the common factor of x.

16. anonymous

Great. Thanks for your help :)

17. anonymous

No probs.