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UnkleRhaukus

  • 4 years ago

\[{dy \over dx} + {2x-y-4 \over x-y+5}=0\]

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  1. JamesJ
    • 4 years ago
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    You'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 + u-9)/u = 0 This isn't quite simple enough, but you get the idea.

  2. UnkleRhaukus
    • 4 years ago
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    yeah ill show you what i got up to

  3. UnkleRhaukus
    • 4 years ago
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    \[{dy \over dx} +{2x-y-4 \over 2y-x+5}=0\]\[{dy \over dx} = {y-2x+4 \over 2y-x+5}\]\[{dY \over dX} = {(Y+p)-2(X+q)+4 \over 2(Y+p)-(X+q)+5}\]\[{dY \over dX} = {Y-2X+(p-2q+4) \over 2Y-X+(2p-q+5)}\]

  4. UnkleRhaukus
    • 4 years ago
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    to find p and q we use \[p-2q+4=0\]\[2p-q+5=0\] \[p=-2\]\[q=1\]

  5. JamesJ
    • 4 years ago
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    Right. And now you can write the new equation as a homogeneous equation.

  6. UnkleRhaukus
    • 4 years ago
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    \[{dY \over dX}={Y-2X \over 2Y-X}\]\[{dY \over dX}={Y/X-2 \over 2Y/X-1}\]\[X{dV \over dX}+V={V-2 \over 2V-1}\]\[X{dV \over dX}={-V^2+2V-2 \over 2V-1}\]\[{ 2V-1 \over V^2-V+1} {dV}=-2{dX \over X}\]\[\int{ 2V-1 \over V^2-V+1} {dV}=-2 \int {dX \over X}\]\[ln(V^2-V+1)=-2lnX+c\]\[V^2-V+1=kX^{-2}\]\[Y^2-YX+X^2=k\]\[(y+2)^2-(y+2) \times (x-1)+(x-1)-k=0\]\[x^2+y^2-xy-4x+5y-k=0\]

  7. UnkleRhaukus
    • 4 years ago
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    is there quicker way?

  8. JamesJ
    • 4 years ago
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    I don't think so.

  9. JamesJ
    • 4 years ago
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    What is that btw ... a skewed ellipse I think

  10. UnkleRhaukus
    • 4 years ago
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    yes a skewed ellipse something like |dw:1327457371333:dw| how can you tell that , ?

  11. JamesJ
    • 4 years ago
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    You can see how might factor is as something like a^2(x-ky)^2 + b^2y^2 = 1

  12. UnkleRhaukus
    • 4 years ago
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    kinda,

  13. UnkleRhaukus
    • 4 years ago
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    I have some similar questions that i can not seem to solve for p,q their seems be no solution \[{dy \over dx} ={x-y+2 \over x-y+1}\]

  14. JamesJ
    • 4 years ago
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    For this one, I'd set u = x - y + 1 u' = 1 -y' ode => 1 - u' = (u+1)/u This equation is separable.

  15. JamesJ
    • 4 years ago
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    In fact, it's u' = -1/u

  16. UnkleRhaukus
    • 4 years ago
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    \[u=-\int{1 \over u}du\]\[u=-ln|u|+c\]\[e^u=-ln(u)+c\]\[e^u=k/u\]\[ue^u=k\]\[(x-y+1)e^{x-y+1}-k=0\] i went wrong somewhere

  17. JamesJ
    • 4 years ago
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    must be tired. This one's easy \[ u du = -dx \]

  18. UnkleRhaukus
    • 4 years ago
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    \[u^2=-2x+c\]\[u^2+2x-c=0\]\[(x−y+1)^2+2x-c=0\] √ got there thank you and yes i am tired

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