Question

Can someone help me solve this Differential equation? dy+lnxy dx=(4x+lny)dx

Answer 1

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

Answer 2

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\[\frac{dy}{dx}=4x-\ln x \rightarrow y=2x^2-x \ln x +x+c\]

Answer 3

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

Answer 4

Do you know how to integrate ln(x)?

Answer 5

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

Answer 6

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

Answer 7

If so, yes.

Answer 8

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

Answer 9

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 \[2x^2-x \ln x + x +c\]

Answer 10

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

Answer 11

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

Answer 12

Yep

Answer 13

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

Answer 14

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

Answer 15

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.

Answer 16

Great. Thanks for your help :)

Answer 17

No probs.