A community for students.
Here's the question you clicked on:
 0 viewing
UnkleRhaukus
 4 years ago
\[{dy \over dx} + {2xy4 \over xy+5}=0\]
UnkleRhaukus
 4 years ago
\[{dy \over dx} + {2xy4 \over xy+5}=0\]

This Question is Closed

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2You'll want to change variables to help you get to a simpler equation. Something like u = x  y + 5 Then u' = 1  y' and your equation becomes 1  u' + (x + u9)/u = 0 This isn't quite simple enough, but you get the idea.

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0yeah ill show you what i got up to

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0\[{dy \over dx} +{2xy4 \over 2yx+5}=0\]\[{dy \over dx} = {y2x+4 \over 2yx+5}\]\[{dY \over dX} = {(Y+p)2(X+q)+4 \over 2(Y+p)(X+q)+5}\]\[{dY \over dX} = {Y2X+(p2q+4) \over 2YX+(2pq+5)}\]

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0to find p and q we use \[p2q+4=0\]\[2pq+5=0\] \[p=2\]\[q=1\]

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2Right. And now you can write the new equation as a homogeneous equation.

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0\[{dY \over dX}={Y2X \over 2YX}\]\[{dY \over dX}={Y/X2 \over 2Y/X1}\]\[X{dV \over dX}+V={V2 \over 2V1}\]\[X{dV \over dX}={V^2+2V2 \over 2V1}\]\[{ 2V1 \over V^2V+1} {dV}=2{dX \over X}\]\[\int{ 2V1 \over V^2V+1} {dV}=2 \int {dX \over X}\]\[ln(V^2V+1)=2lnX+c\]\[V^2V+1=kX^{2}\]\[Y^2YX+X^2=k\]\[(y+2)^2(y+2) \times (x1)+(x1)k=0\]\[x^2+y^2xy4x+5yk=0\]

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0is there quicker way?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2What is that btw ... a skewed ellipse I think

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0yes a skewed ellipse something like dw:1327457371333:dw how can you tell that , ?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2You can see how might factor is as something like a^2(xky)^2 + b^2y^2 = 1

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0I have some similar questions that i can not seem to solve for p,q their seems be no solution \[{dy \over dx} ={xy+2 \over xy+1}\]

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2For this one, I'd set u = x  y + 1 u' = 1 y' ode => 1  u' = (u+1)/u This equation is separable.

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0\[u=\int{1 \over u}du\]\[u=lnu+c\]\[e^u=ln(u)+c\]\[e^u=k/u\]\[ue^u=k\]\[(xy+1)e^{xy+1}k=0\] i went wrong somewhere

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2must be tired. This one's easy \[ u du = dx \]

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0\[u^2=2x+c\]\[u^2+2xc=0\]\[(x−y+1)^2+2xc=0\] √ got there thank you and yes i am tired
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.