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gerryliyana

  • 2 years ago

How to find Solution to differential equation (x - 4) y^{4} dx - x^{3} (y^{2} - 3 ) dy = 0

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  1. gerryliyana
    • 2 years ago
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    How to find Solution to not exact differential equation \[(x - 4) y^{4} dx - x^{3} (y^{2} - 3 ) dy = 0\]

  2. henpen
    • 2 years ago
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    Isn't \[\int\limits \frac{x-4}{x^3}dx=\int\limits \frac{y^2-3}{y^4}dy\] enough?

  3. henpen
    • 2 years ago
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    It's probably not, I'm still pants at DE

  4. henpen
    • 2 years ago
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    Anyway: don't ask me HOW to integrate that (use Wolfram integrator if you just want the answer), it's probably some counter-intuitive trig substitution.

  5. gerryliyana
    • 2 years ago
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    i got \[- \frac{ 1 }{ x } + \frac{ 2 }{ x^{2} }-\frac{ 1 }{ y } + \frac{ 1 }{ y^{3} } = C\] am i right?

  6. henpen
    • 2 years ago
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    http://integrals.wolfram.com/index.jsp?expr=%28x-4%29%2Fx%5E3&random=false http://integrals.wolfram.com/index.jsp?expr=%28x%5E2-3%29%2Fx%5E4&random=false So your signs are wrong for either the x or y terms but it's fine other than that.

  7. niksva
    • 2 years ago
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    @gerryliyana in this type of differntial eqution, we use separation of variable technique as we can see from this differential equation, variables can be separated easily

  8. niksva
    • 2 years ago
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    @ gerryliyana i m getting -1/x + 2/x^2 + 1/ y^2 - 1 /y^3 = C where C is constant

  9. niksva
    • 2 years ago
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    @gerryliyana right????

  10. gerryliyana
    • 2 years ago
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    ok @niksva but i got -1/x + 2/x^2 + 1/ y - 1 /y^3 = C where C is constant

  11. niksva
    • 2 years ago
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    omg my mistake it is 1/y

  12. gerryliyana
    • 2 years ago
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    no problem :)

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