A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
Differential Equations(Integration Factor):
show that (x+y)^(2) is the integration factor of [x^(2)+2xyy^(2)]dx + [y^(2)+2xyx^(2)]dy = 0
Need Help, thank you very much.
anonymous
 one year ago
Differential Equations(Integration Factor): show that (x+y)^(2) is the integration factor of [x^(2)+2xyy^(2)]dx + [y^(2)+2xyx^(2)]dy = 0 Need Help, thank you very much.

This Question is Closed

danica518
 one year ago
Best ResponseYou've already chosen the best response.1http://38.media.tumblr.com/1741e6ccb28233179cc63040cdf09524/tumblr_n2hhznE3Ub1rfduvxo1_250.gif

danica518
 one year ago
Best ResponseYou've already chosen the best response.1is this the given equation exact

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think it is nonexact since the integration factor is given :)

danica518
 one year ago
Best ResponseYou've already chosen the best response.1take df/dx = x^(2)+2xyy^(2) df/dy=y^(2)+2xyx^(2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0getting the partial derivatives results to M'(y)=2x2y;N'(x)=2y2x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which can be summarized to M'(y)=N'(x), but i want to know how did (x+y)^(2) is solved

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh.... so, how did the integration factor became (x+y)^(2)? that is what i can't understand ^^;

freckles
 one year ago
Best ResponseYou've already chosen the best response.2well have you tried dividing both sides by (x+y)^2 and checking if M_y=N_x?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2that is all you need to show that it is the integrating factor you need

danica518
 one year ago
Best ResponseYou've already chosen the best response.1you gotta just show the integration is true here, not how its found,

danica518
 one year ago
Best ResponseYou've already chosen the best response.1im curious how ud go about finding that integration factor here too

danica518
 one year ago
Best ResponseYou've already chosen the best response.1it might from like a sturmlouiville form t

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\text{ if } \frac{1}{N}[M_yN_x]=h(x) \text{ then the integrating factor is } e^{\int\limits h(x) dx} \\ \text{ if } \frac{1}{M}[N_xM_y]=k(y) \text{ then the integrating factor is } e^{\int\limits k(y) dy}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2where you have the differential equation:\[Mdx+Ndy=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@freckles \[M=12[\frac{ \frac{ y }{ x } }{ 1+\frac{ y }{ x } }]^{2} ; N=12[\frac{1}{ 1+\frac{ y }{ x } }]^{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that's what i got when i divide them with (x+y)^(2) ^^;

danica518
 one year ago
Best ResponseYou've already chosen the best response.1show the newequation u get after multiplying by integration factor is still exact

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[M=x^2+2xyy^2 \\ M=x^2+2xy+y^22y^2 \\ M=(x+y)^22y^2 \\ \frac{M}{(x+y)^2}=1\frac{2y^2}{(x+y)^2} \\ N=y^2+2xyx^2 \\ N=y^2+2xy+x^22x^2 \\ N=(x+y)^22x^2 \\ \frac{N}{(x+y)^2}=1\frac{2x^2}{(x+y)^2}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\text{ new } M=1\frac{2y^2}{(x+y)^2} \\ \text{ new} N=1\frac{2x^2}{(x+y)^2}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2by the way those two things I mentioned won't work since you don't get a function of x or a function of y you get a function of x and y for either try but anyways all you need to do is show new M_y=new N_x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the point of an integration factor is to turn an 'inexact' differential \(P\, dx+M\, dy\) into an 'exact' or total one \(d\Phi=\mu P\, dx+\mu M\, dy\), which is equivalent to solving the following PDE: $$\frac{\partial}{\partial y}[\mu P]=\frac{\partial}{\partial x}[\mu M]$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since we want $$\mu P=\frac{\partial\Phi}{\partial x},\mu M=\frac{\partial\Phi}{\partial y}$$ and \(\dfrac{\partial^2\Phi}{\partial y\,\partial x}=\dfrac{\partial^2\Phi}{\partial x\,\partial y}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0anyways, the standard solution presumes \(\mu\) is a function of a single variable, either \(\mu(x)\) or \(\mu(y)\) so $$\frac{\partial}{\partial y}[\mu P]=\frac{\partial\mu}{\partial y}P+\mu\frac{\partial P}{\partial y}\\\frac{\partial}{\partial x}[\mu M]=\frac{\partial\mu}{\partial x}M+\mu\frac{\partial M}{\partial x}$$ so $$\frac{\partial\mu}{\partial y}P+\mu\frac{\partial P}{\partial y}=\frac{\partial\mu}{\partial x}M+\mu\frac{\partial M}{\partial x}\\P\frac{\partial\mu}{\partial y}M\frac{\partial\mu}{\partial x}+\left(\frac{\partial P}{\partial y}\frac{\partial M}{\partial x}\right)\mu=0$$ which then reduces to a linear firstorder ODE if we choose either \(\mu_y=\frac{d\mu}{dy},\mu_x=0\) or \(\mu_x=\frac{d\mu}{dx},\mu_y=0\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for the help everyone :) i haven't finished solving yet but i got the concept. Thanks again.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.