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Math Help - Differential Equation Substitution help.

  1. #1
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    Differential Equation Substitution help.

    Problem goes as follows :

    (y^2 + x^2)dx + (2xy)dy = 0

    Now I can clearly see it is homogeneous but my problem starts once I get it in derivative form. Once in derivative form, I subbed u = y/x , y = ux , dy/dx = (du/dx) x + u.


    Once I subbed it all in, I was left with

    u + du/dx = - (1 + u^2) / (2u)

    Can someone help me proceed from here?
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  2. #2
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    Quote Originally Posted by JonathanEyoon View Post
    Problem goes as follows :

    (y^2 + x^2)dx + (2xy)dy = 0

    Now I can clearly see it is homogeneous but my problem starts once I get it in derivative form. Once in derivative form, I subbed u = y/x , y = ux , dy/dx = (du/dx) x + u.


    Once I subbed it all in, I was left with

    u + du/dx = - (1 + u^2) / (2u)

    Can someone help me proceed from here?
    Sure, move the u from the left to the right hand side and simplfiy. Now you can separate.
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  3. #3
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    Quote Originally Posted by danny arrigo View Post
    Sure, move the u from the left to the right hand side and simplfiy. Now you can separate.


    But that would net me two terms on the right side which isn't separable. ?
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  4. #4
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    LOL nevermind. Can't believe I forgot to simplify . Thanks dood
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  5. #5
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    Quote Originally Posted by JonathanEyoon View Post
    But that would net me two terms on the right side which isn't separable. ?
    But all the terms are in the variable v which you can simplify.

    Edit - You got it!
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  6. #6
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    Hate to ask of this but can anyone give me a final answer for me? This problem doesn't have an answer in the back of my textbook . I got it down to

    (-1/3)(ln(3(y^2/x^2) + 1) = x + C
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  7. #7
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    Quote Originally Posted by JonathanEyoon View Post
    Problem goes as follows :

    (y^2 + x^2)dx + (2xy)dy = 0

    Now I can clearly see it is homogeneous but my problem starts once I get it in derivative form. Once in derivative form, I subbed u = y/x , y = ux , dy/dx = (du/dx) x + u.


    Once I subbed it all in, I was left with

    u + du/dx = - (1 + u^2) / (2u)

    Can someone help me proceed from here?
    Another way (a little quicker) is to re-write

    <br />
(y^2 + x^2)dx + (2xy)dy = 0<br />

    as

    d(x \,y^2) + x^2 dx = 0

    which integrates giving

    x \,y^2 + \frac{1}{3} x^3 = c
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  8. #8
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    Quote Originally Posted by danny arrigo View Post
    Another way (a little quicker) is to re-write

    <br />
(y^2 + x^2)dx + (2xy)dy = 0<br />

    as

    d(x \,y^2) + x^2 dx = 0

    which integrates giving

    x \,y^2 + \frac{1}{3} x^3 = c

    Could you show me how you simplified that? Also the way I did it, I was left with what I gave on the post above yours. I'm not sure where i'm going wrong here. Any help is appreciated.
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  9. #9
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    Quote Originally Posted by JonathanEyoon View Post
    Problem goes as follows :

    (y^2 + x^2)dx + (2xy)dy = 0

    Now I can clearly see it is homogeneous but my problem starts once I get it in derivative form. Once in derivative form, I subbed u = y/x , y = ux , dy/dx = (du/dx) x + u.


    Once I subbed it all in, I was left with

    u + du/dx = - (1 + u^2) / (2u)

    Can someone help me proceed from here?
    Let's pick it up from here. When we isolate u' we get

    \frac{du}{dx} =- \frac{1+3u^2}{2u}.

    Separating gives

    \frac{2u}{1+3u^2}\,du = - \frac{dx}{x}

    integrating

    \frac{1}{3} \ln |\, 1 + 3u^2\, | = - \ln |\,x\,| + \frac{1}{3} \ln c

    multiply by 3, moving the \ln |x| to the RHS and exponentiating

    (1+3u^2)x^3 = c

    which after substituting u = \frac{y}{x} and simplifying gives the answer

    x^3 + 3xy^2 = c.

    Hope it helps.
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  10. #10
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    That helped more than you can imagine . I really appreciate it!
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  11. #11
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    Ok sorry but I have no idea where you got the


    + 1/3 lnC

    Shouldn't it be just lnX + C?
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  12. #12
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    Quote Originally Posted by JonathanEyoon View Post
    Ok sorry but I have no idea where you got the


    + 1/3 lnC

    Shouldn't it be just lnX + C?
    Sure. So integrating

    \frac{1}{3} \ln |\, 1 + 3u^2\, | = - \ln |\,x\,| + c

    multiply by 3, moving the \ln |x| to the RHS and exponentiating

    (1+3u^2)x^3 = e^{3c}

    which after substituting u = \frac{y}{x} and simplifying gives the answer

    x^3 + 3xy^2 = e^{3c}.

    Since c is a constant then  e^{3c} which we'll call C

    (1+3u^2)x^3 = C same answer. So why did I use  \frac{1}{3} \ln c. I saw that I was going to multiply by 3 and exponentiate.
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  13. #13
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    I hate to ask this but could you show me in more detailed steps or least tell me in more detail how you went from after integrating and then getting

    (1 + 3u^2)x^3.


    I get that e^3c is just another constant so that'll equal C later but i'm at an absolute loss as to how you got what I typed above. If you multiply both sides by 3, you should end up with

    ln(1+3u^2) = -3lnx + 3c

    If you take e to the both sides, you should get

    (1 + 3u^2) = - 1/x^3 + e^3c

    now I can get this far... and I just can't see how you're able to get

    (1 + 3u^2)x^3 on the left hand side and e^3c on the right hand side.

    I mean supposing you DID multiply both sides by x^3, THEN I can see you getting that for the left side but on the right side you'll end up with -1 + (x^3)(e^3c)

    .......If you don't mind could you show it to me in clearer steps? I've been racking my brains for over a day on this problem and I can't seem to get it to how you're able to get it to. THanks.
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  14. #14
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    Quote Originally Posted by JonathanEyoon View Post
    I hate to ask this but could you show me in more detailed steps or least tell me in more detail how you went from after integrating and then getting

    (1 + 3u^2)x^3.


    I get that e^3c is just another constant so that'll equal C later but i'm at an absolute loss as to how you got what I typed above. If you multiply both sides by 3, you should end up with

    ln(1+3u^2) = -3lnx + 3c

    If you take e to the both sides, you should get

    (1 + 3u^2) = - 1/x^3 + e^3c Mr F says: No. See below.

    now I can get this far... and I just can't see how you're able to get

    (1 + 3u^2)x^3 on the left hand side and e^3c on the right hand side.

    I mean supposing you DID multiply both sides by x^3, THEN I can see you getting that for the left side but on the right side you'll end up with -1 + (x^3)(e^3c)

    .......If you don't mind could you show it to me in clearer steps? I've been racking my brains for over a day on this problem and I can't seem to get it to how you're able to get it to. THanks.
    e^{\ln |x|^{-3} + 3C} = e^{\ln |x|^{-3}} \cdot e^{3C} = \frac{1}{x^3} \cdot e^{3C} NOT \frac{1}{x^3} {\color{red}+} e^{3C}.
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