Homogeneous Differential Equation

  • Thread starter Dr.Doom
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  • #1
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Homework Statement



Find the general solution of the differential equation:

(x+y)y'=x-y


Homework Equations



I want to solve this as a homogeneous differential equation, so our equations are:
v=[itex]\frac{y}{x}[/itex], y=vx, [itex]\frac{dy}{dx}[/itex]=v+x[itex]\frac{dv}{dx}[/itex]


The Attempt at a Solution



I need to get this into the form [itex]\frac{dy}{dx}[/itex]=F([itex]\frac{y}{x}[/itex]), so I rewrite it as y'=[itex]\frac{x-y}{x+y}[/itex]. Dividing by x I get, y'=[itex]\frac{1-\frac{y}{x}}{1+\frac{y}{x}}[/itex]. From here, I substitute to get v+x[itex]\frac{dv}{dx}[/itex]=[itex]\frac{1-v}{1+v}[/itex]. When looking at the solution manual, however, it says that it should be in the form x(v+1)v'=-(v2+2v-1) before I integrate. I don't see how i can get it into this form. Also, it gives the answer as y2+2xy-x2=C. I can correctly solve this differential equation using different methods, but I would really like to know how to solve this using the homogeneous method. I would really appreciate any help with this. Thanks!
 

Answers and Replies

  • #2
hunt_mat
Homework Helper
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You are almost there...

You have to compute:
[tex]
\frac{1-v}{1+v}-v=x\frac{dv}{dx}
[/tex]
This will be a separable equation which is solved in the usual fashion.
 
  • #3
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you just have to take subtract v on both sides of the equation you got man......(xdv/dx=(1-v)/(1+v) - v..
 
  • #4
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Ok, so I rewrite it as [itex]\frac{1}{x}[/itex]dx=[itex]\frac{1+v}{1-v}[/itex]-[itex]\frac{1}{v}[/itex]dv. Do I need to do an additional substitution on [itex]\frac{1+v}{1-v}[/itex]?
 
  • #5
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you solved it wrong on right side... you will get dv/{(1-v)/(1+V) - v}....which comes out to be (1+v)/(1-v^2-2v)...just substitute 1-v^2-2v as t....
 
  • #6
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I'm having trouble seeing how it comes out to (1+v)/(1-v2-2v)
 
  • #7
hunt_mat
Homework Helper
1,745
26
Do the addition as I said to get:
[tex]
\frac{1-v}{1+v}-v=\frac{1-v-v(1+v)}{1+v}=\frac{1-v-v-v^{2}}{1+v}=\frac{1-2v-v^{2}}{1+v}
[/tex]
 

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