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I I need help with differential equation(3x^2-2y^2)(dy/dx)=2xy

  1. Aug 15, 2016 #1
    I have been stuck in this problem for two days now. I am starting DE this semester and I want to move ahead. so for this problem I attempted to use x=yv, then v=x/y. so I move dy/dx to the other side of the equation and divide by 2xy both sides, which leads to (3x^2-2y^2)/(2xy)=dx/dy. then if x=vy, then (3x^2-2y^2)/(2xy)=(dv/dy)y+v. then, (3/2)(x/y) -(y/x) -v=(dv/dy)y. then [(3/2)v-(1/v)-v]=(dv/dy)y. There will be a common demonimator of 2v on the right of the equation, leading to (3v^2-2-2v^2)/(2v)=(dv/dy)y. then after algebra, dy/y=(2v/(v^2-2))dv. Integrating leads to y=v^2-2, which is y=(x^2/y^2)-2. but in my book the answer comes with an additional factor of two in the answer. so the right answer should be y=(x^2/2y^2)-2. I found that if divide by only xy on both side of the equations, and later dividing by two when integrating, it will lead me to that right answer, but I don't understand why. there is a similar problem , with the expection that there is no 2 in xy. and I got that one right, but not this one. help please!!
     
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  3. Aug 15, 2016 #2

    haruspex

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    What about the constant of integration?
     
  4. Aug 15, 2016 #3
    if v^2-2=u, then du=2v, which cancels the 2v that is above of the denominator right? help!! how would you do it using the method i used?
     
  5. Aug 15, 2016 #4

    haruspex

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    2dv.
    But you did not answer my question. Go through the integration step and make sure to include a constant of integration. If you still do not see your error, please post all your steps in detail.
     
  6. Aug 15, 2016 #5
    I dont see my error!! i will post my steps: (3x^2-2y^2)(dy/dx)=2xy. (3x^2-2y^2)/(2xy)=dx/dy. (3x^2/2xy)-(2y^2/2xy)=dx/dy. 3(x/y)/2-(y/x)=dx/dy. x=yv, v=x/y.
    (3/2)v-(1/v)=(dv/dy)y+v. (3/2)v-(1/v)-v=(dv/dy)y. find lcd for left side of equation. lcd is 2v, so (3v^2-2-2v^2)/(2v)=(dv/dy)y. make dv on one side and dy on the other side, so (2v/(v^2-2))dv=dy/y. integral of (2v/(v^2-2))dv: u=v^2-2, du=2vdv. dv=du/2v, the 2v cancels out, leading to a resulting integral of ln|v^2-2|=ln|y|+c. multiply both sides by e leads to, y=(x^2/y^2)-2+c. the values are y=-1 and x=0. so c=y-(x^2/y^2)+2. c=-1+2=1. y=(x^2/y^2)-2+1. y=(x^2/y^2)-1. y=(x^2-y^2)/(y^2).
    y^3=x^2-y^2. y^3+y^2=x^2. y^2(y+1)=x^2. but the answer is 2y^2(y+1)=x^2. my question would be the factor of two on the right side. help!!!
     
  7. Aug 15, 2016 #6

    haruspex

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    No, you do not multiply by e to remove the ln()s. What do you do?
    That is wrong.
     
  8. Aug 15, 2016 #7
    I thought that it was possible to cancel the natural log by e^x since it is the opposite of the natural log. so will the constant c also be e^c? and I solve for c? help me already!! jaja
     
  9. Aug 15, 2016 #8
    oh I got it!!! so ln|y|=ln|c((x^2/y^2)-2). x=0. y=-1. I use e^x on both side of the equation. so c=1/2. then y=1/2((x^2/y^2)-2). then y=(x^2-2y^2)/(2y^2). finally 2y^3+2y^2=x^2. correct? I have never worked a problem were I have to found what c is, so maybe that was my problem?
     
  10. Aug 16, 2016 #9

    haruspex

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    Where ln(|c|) is the constant of integration, yes.
     
  11. Aug 16, 2016 #10
    thank you very much :P
     
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