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RolyPoly

  • 2 years ago

\[(x+xy^2)dx + e^{x^2}ydy =0\]

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  1. RolyPoly
    • 2 years ago
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    \[(x+xy^2)dx + e^{x^2}ydy=0\]\[x(1+y^2)dx =- e^{x^2}ydy\]\[-\frac{x}{e^{x^2}}dx=\frac{y}{1+y^2}dy\]\[-\frac{1}{2}e^{-x^2}d(x^2)=\frac{1}{2(1+y^2)}d(y^2)\]\[\frac{1}{2}e^{-x^2}+C=\frac{1}{2}\ln |1+y^2|\]\[e^{-x^2}+C=\ln |1+y^2|\]Does it look fine?

  2. hartnn
    • 2 years ago
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    it looks correct to me.

  3. cnknd
    • 2 years ago
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    i dont see anything wrong here

  4. RolyPoly
    • 2 years ago
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    Oh!! Thanks!! Should I further simplify it and make y as the subject or just leave it here?

  5. hartnn
    • 2 years ago
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    no, point.... that will just complicate the equation. exponential of exponential.....

  6. RolyPoly
    • 2 years ago
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    Okay, thanks! :)

  7. hartnn
    • 2 years ago
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    welcome ^_^

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