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anonymous

  • 5 years ago

differentiate:xy'+2y=x^2

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  1. anonymous
    • 5 years ago
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    I have one solution: c/x^2+x^2/4; but confusion, could any one help me

  2. amistre64
    • 5 years ago
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    define for me "differentiate" as it applies to this problem please...

  3. amistre64
    • 5 years ago
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    ...ordinary differential equation....right?

  4. anonymous
    • 5 years ago
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    yes

  5. amistre64
    • 5 years ago
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    if I recall correctly, that means your trying to find the original function that this was derived from correct?

  6. anonymous
    • 5 years ago
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    yeah

  7. amistre64
    • 5 years ago
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    the only method I remember off hand is the seperation of variables.... have you tried that yet?

  8. anonymous
    • 5 years ago
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    i think seperation variable is not correct for this equation

  9. amistre64
    • 5 years ago
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    youre probably right...step me through what youve done already

  10. anonymous
    • 5 years ago
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    i apply bernoulli's equa.

  11. amistre64
    • 5 years ago
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    y' + P(x)y = Q(x)y^n.. that one?

  12. anonymous
    • 5 years ago
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    y'+p(x)y=r(x)

  13. amistre64
    • 5 years ago
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    hmmm.... I havent had much practice with ode's .... maybe someone smarter will come along :)

  14. anonymous
    • 5 years ago
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    You can use the method: multiplying with an integrating factor. divide everything by x, so you'll have y'+2/x y=x. The integrating factor will then be: e^{2\int 1/x}=e^{2lnx}=e^{ln(x^2)}=x^2.

  15. anonymous
    • 5 years ago
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    Multiply both sides by the integrating factor x^2, then the left side will be (y*x^2)'. The right side is x^3. Then you integrate both sides, and gets y*x^2={1/4} *x^4+C Divide both sides of the equation by x^2 and you have the answer :) If you're not familiar with the method and wants a further explanation, just tell me...

  16. anonymous
    • 5 years ago
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    c/x^2+x^2/4 i have already mention above

  17. anonymous
    • 5 years ago
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    is it correct?

  18. anonymous
    • 5 years ago
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    Yes, it is correct. y=... , youre right. I only looked too much at the latter posts instead of where your answer was, I'm sorry for that..

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