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RolyPoly

  • 3 years ago

Solve: \[(2xy)dx + (y^2-3x^2)dy=0\]

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  1. sirm3d
    • 3 years ago
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    \[\frac{ \partial }{ \partial y }(2xy)=2x,\frac{ \partial }{ \partial x }(y^2-3x^2)=-6x\]\[\frac{ -(\frac{ \partial M }{ \partial y } - \frac{ \partial N }{ \partial x })}{ 2xy }=-\frac{ 4 }{ y }\]the integrating factor that will make the DE exact is \[\large e^{\int\limits_{}^{}\frac{ -4 }{ y }dy}=\frac{ 1 }{ y^4 }\]

  2. sirm3d
    • 3 years ago
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    if you have not reached this part of the course, an alternate solution is \[\large x=vy\] and \[\large dx = vdy + ydv\] where the original DE is now separable in v and y.

  3. sirm3d
    • 3 years ago
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    have you solved this problem @RolyPoly ?

  4. RolyPoly
    • 3 years ago
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    I'm on the way solving it.. Sorry, please give me some more time...

  5. RolyPoly
    • 3 years ago
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    I'm not quite sure if I have learnt this way before. I have no memory of seeing this, but it appears in my book..

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  6. sirm3d
    • 3 years ago
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    homogeneous function, doesn't it ring a bell?

  7. RolyPoly
    • 3 years ago
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    Nope :(

  8. sirm3d
    • 3 years ago
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    my comment on your attached file: \[\huge F=\frac{ x^2 }{ y^3 }+g_1(y)\] NOT \[\huge F_x\] and \[\huge F=-\frac{1}{y}+\frac{x^2}{y^3}+g_2(x)\] instead of \[\huge F_y\]

  9. RolyPoly
    • 3 years ago
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    :O That's how my teacher writes... \(F_x\) means integrating w.r.t. x, I think :S

  10. sirm3d
    • 3 years ago
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    about that alternate solution \[\large \deg(2xy)=2;\deg(y^2)=2;\deg(-3x^2)=2\] the degrees of the terms are identical. a substitution \[\large y = ux \text{ or }x=vy\] will transform the DE that is solvable by separation

  11. sirm3d
    • 3 years ago
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    taking \[x=vy \text{ and } dx=vdy+ydv\]\[\large 2(vy)y(vdy+ydv)+(y^2-v^2y^2)dy=0\]

  12. RolyPoly
    • 3 years ago
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    I suppose it is separable, but somehow, I can't separate it. I guess I have to check my calculations again first :/

  13. sirm3d
    • 3 years ago
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    collecting terms of dy and dv, \[\large (2v^2y^2+y^2-v^2y^2)dy+2vy^3dv=0\]or\[\large y^2(v^2+1)dy+2vy^3dv=0\]divide by \[\large y^3(v^2+1)\] and the DE becomes \[\large \frac{dy}{y}+\frac{2vdv}{v^2+1}=0\]

  14. RolyPoly
    • 3 years ago
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    How come... you can get it but I can't... :'(

  15. sirm3d
    • 3 years ago
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    expand after replacing x by vy and dx by vdy + ydv collect terms of dy and dv, simplify and factor.

  16. RolyPoly
    • 3 years ago
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    I tried w = y/x but it doesn't seem to be working for me :(

  17. RolyPoly
    • 3 years ago
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    Sorry, I got to go now.. I'd keep trying until I got the answer. Thanks for your help!!

  18. RolyPoly
    • 3 years ago
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    @sirm3d You're right about the F and Fx thing! Sorry for my earlier mistakes!!! When I tried w=y/x, I got something quite ugly :S

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