First-Order Linear ODE help?

First-Order Linear ODE help???

This is my first post here. I'm still getting used to LATEX syntax so please forgive any mistakes. My question is on a simple differential equation... it doesn't appear to be exact or homogeneous... where do I start?

thanks in advance.

DD86

 

[tex]y'-4xy+2yln(y)/x=0[/tex]

[tex]M(x,y)= -4xy \quad and \quad N(x,y)=2yln(y)/x[/tex]

err.. I'm taking Calc 3 and diff eq at the same time so I hope my partials are right...

[tex]M_{y}=-4x \quad and \quad N_{x}=0[/tex]

IntegFactor is defined as [tex]I(x) = e^{\int (M_{y} - N_{x})/N dx} [/tex] according to my notes. yeah I'm still stuck. Could someone verify the partial derivatives though?

EDIT: I now have [tex]y'+2y(-2x+ln(y)/x)=0[/tex] which looks nicer but im still at a loss for what to do next...

What integrating factor am I looking for?

 

thanks for your input saltdog. My prof seems to like asking questions where you need to recognize "chain-ruled" things. For example, I have to get used to identifying a product of two terms as the derivative of some compossite function... not cool hehe

 

smart trick Tide!

here's my mess of a solution:

[tex]xy'+4x^2y-2ylny=0[/tex]
[tex]y'-4xy=(-2ylny)/x[/tex]

[tex]Let \quad y = e^{g(x)} \implies y'=g'e^{g(x)}[/tex]

[tex]g'e^{g(x)}-4xe^{g(x)}= \frac {-2g(x)e^{g(x)}}{x}[/tex]

divide out e^g(x) and rearrange

[tex]g'+\frac {2}{x}g=4x, Let \quad P(x) = 2/x \quad and \quad Q(x)=4x[/tex]

[tex]g(x) = e^{- \int P(x) dx}[\int Q(x)e^{\int P(x) dx} dx + C][/tex]

[tex]\frac {1}{x^2}[x^4+C] \implies g(x)=x^2 + \frac {C}{x^2}[/tex]

QUESTION: am I allowed to call the "c/x^2" term just another constant?

back-sub [tex]y(x) = e^{g(x)} = e^{(x^2)}*e^{\frac {C}{x^2}}[/tex]


Thanks for all of your help.