First-Order Linear ODE help?

[decemberdays86]

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

 

[Tide]

HINT: Look for an integrating factor.

 

[decemberdays86]

$$y'-4xy+2yln(y)/x=0$$

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

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

$$M_{y}=-4x \quad and \quad N_{x}=0$$

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

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

What integrating factor am I looking for?

 

[saltydog]

December, need to put it into the form:

$$Mdx+Ndy=0$$

Thus for:

$$y^{'}-4xy+\frac{2yln(y)}{x}=0$$

we'd have:

$$(2yln(y)-4x^2y)dx+xdy=0$$

But that's not happening for me either. Can't see a way to solve it.

Tide . . . how about another hint? Somebody else too is ok.

Edit: Oh yea, it's not linear because of the ln(y) term.

 

[Agnostic]

DecemberDays86, This really doesn't have to do with anything, but, do you post on bodybuilding.com also?

 

[decemberdays86]

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

 

[Tide]

How about this?

Divide both sides of your equation by y, note that y'/y = (ln y)', multiply both sides by x and finally arrive at

$$\frac {d}{dx} \left( x^2 \ln y \right) =4 x^3$$

then see where you can go from there!

 

[saltydog]

How about this?

Divide both sides of your equation by y, note that y'/y = (ln y)', multiply both sides by x and finally arrive at

$$\frac {d}{dx} \left( x^2 \ln y \right) =4 x^3$$

then see where you can go from there!

Ok, I get it now. Thanks Tide. You too December . . . PF rocks.

 

[decemberdays86]

smart trick Tide!

here's my mess of a solution:

$$xy'+4x^2y-2ylny=0$$
$$y'-4xy=(-2ylny)/x$$

$$Let \quad y = e^{g(x)} \implies y'=g'e^{g(x)}$$

$$g'e^{g(x)}-4xe^{g(x)}= \frac {-2g(x)e^{g(x)}}{x}$$

divide out e^g(x) and rearrange

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

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

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

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

back-sub $$y(x) = e^{g(x)} = e^{(x^2)}*e^{\frac {C}{x^2}}$$


Thanks for all of your help.