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RolyPoly

  • 3 years ago

\[\frac{dy}{dx} = \frac{x}{y^2\sqrt{1+x}}\]

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  1. RolyPoly
    • 3 years ago
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    \[\frac{dy}{dx} = \frac{x}{y^2\sqrt{1+x}}\]\[y^2dy = \frac{x}{\sqrt{1+x}}dx\]\[\frac{1}{3}y^3= \int \frac{u-1}{u}du = \frac{2}{3} u^{\frac{3}{2}}- 2u^{\frac{1}{2}}+C\]\[y = 2(1+x)^{\frac{3}{2}}-6(1+x)^{\frac{1}{2}}+C)^{\frac{1}{3}}\]

  2. hartnn
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=integral+x%2Fsqrt%281%2Bx%29

  3. RolyPoly
    • 3 years ago
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    Did I do something wrong again?

  4. hartnn
    • 3 years ago
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    i couldn't understand clearly what u did, but it can be simplified to 2/3*(x-2)*sqrt(x+1)

  5. RolyPoly
    • 3 years ago
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    Which part you didn't understand?

  6. hartnn
    • 3 years ago
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    u=?

  7. RolyPoly
    • 3 years ago
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    1+x

  8. hartnn
    • 3 years ago
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    then denominator is sqrt u

  9. RolyPoly
    • 3 years ago
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    OMG!I typed something wrong!! \[\frac{1}{3}y^3= \int \frac{u-1}{\sqrt u}du = \frac{2}{3} u^{\frac{3}{2}}- 2u^{\frac{1}{2}}+C\]

  10. RolyPoly
    • 3 years ago
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    No wonder why you couldn't understand.. That was my mistake :( Is that clear now?

  11. hartnn
    • 3 years ago
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    yeah...u can simplify your final answer by factoring out sqrt(x+1)

  12. hartnn
    • 3 years ago
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    if needed.

  13. RolyPoly
    • 3 years ago
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    That's how the answer in my book looks like :S But thanks!

  14. hartnn
    • 3 years ago
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    ohh.. then no need. you were correct all along. welcome ^_^

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