Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

gerryliyanaBest ResponseYou've already chosen the best response.2
I have: \[(2x^{2} + y) dx + (x^{2} y  x) dy = 0 \]
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.2
did you get it @sirm3d ??
 one year ago

sirm3dBest ResponseYou've already chosen the best response.2
\[\large 2x^2dx+ydx+x^2ydyxdy=0\]\[\large 2x^2dx(xdyydx)+x^2ydy=0\]divide everything by x^2 \[\large2dx\frac{ xdyydx }{ x^2 }+ydy=0\]thesecond term is the differential d(y/x)\[\large2dxd\left(\frac{y}{x}\right)+ydy=0\]integrate\[\large 2x\frac{ y }{ x }+\frac{ y^2 }{ 2 }=\frac{ C }{ 2 }\]\[\large 4x^22y+2xy^2=Cx\]
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.2
I got \[2x + \frac{ 1 }{ 2 } y^{2} \frac{ y }{ x } = C\]
 one year ago

sirm3dBest ResponseYou've already chosen the best response.2
that is also correct. the difference in our answer is the choice of the constant.
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.2
very nice..., Thank You for Checking :)
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.0
dw:1354568766432:dw
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.2
thank u @mahmit2012 i'll post my answer \[(2x^{2} + y) dx + (x^{2} y x) dy = 0\] where \((2x^{2} + y) = M\) and \(x^{2}y  x = N\) \[\frac{ dM }{ dy } = 1\] \[\frac{ dN }{ dy } = 2xy  1\] So this isn't exact differential equations for integration factor: \(P (x) = \frac{ 1 }{ N } \left( \frac{ dM }{ dy }  \frac{ dN }{ dx } \right)\) because it's only f(x) \[P (x) = \frac{ 1 }{ x^{2}y  x } \left[ 1 (2xy 1) \right] = \frac{ 2(1xy) }{ x^{2}yx}= \frac{ 2 }{ x }\] \[\mu(x) = \exp (2 \int\limits_{0}^{x} \frac{ dx }{ x }) = \exp(2 \ln x) = e^{\ln x^{2}} = x^{2} = \frac{ 1 }{ x^{2}}\] \[\mu(x) = \exp(\int\limits_{0}^{x} P(x) dx)\] \[(2+ \frac{ y }{ x^{2} } ) dx + (y  \frac{ 1 }{ x }) dy = 0\] Look! its exact differential equations..., multiply all (1) terms by integration factor, becomes: \[x^{2} (2x^{2} + y) dx + x^{2} (x^{2}y  x) dy = 0\] Then, using grouping method: \[2 dx + \frac{ y }{ x^{2} } dx + y dy  \frac{ dy }{ x } = 0\] \[2dx + y dy +\frac{ y }{ x^{2} } dx  \frac{ dy }{ x } = 0\] \[d(2x) + d(\frac{ 1 }{ 2 }y^{2}) + d(\frac{ y }{ x }) = dC\] \[\int\limits d(2x) + \int\limits d(\frac{ 1 }{ 2 }y^{2}) + \int\limits d(\frac{ y }{ x }) = \int\limits dC\] becomes \[2x + \frac{ 1 }{ 2 } y^{2}  \frac{ y }{ x } =C\]
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.2
I thank all of you guys.., , @sirm3d and @mahmit2012 thank u so much :)
 one year ago
See more questions >>>
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.