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haganmc Group Title

solve diff equation dy/dx =(sec^2(y))/(1+x^2)

  • one year ago
  • one year ago

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  1. lgbasallote Group Title
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    \[\frac{dy}{dx} = \frac{\sec^2 y}{1 + x^2}\] divide both sides by sec^2 y \[\implies \frac{dy}{\sec^2 y dx} = \frac{1}{1+x^2}\] now multiply both sides by dx \[\implies \frac{dy}{\sec^2 y} = \frac{dx}{1+x^2}\] now you can integrate both sides \[\implies \int \frac{dy}{\sec^2 y} = \int \frac{dx}{1+x^2}\] does that help?

    • one year ago
  2. haganmc Group Title
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    yes. the Integration is the problem i'm having trouble with

    • one year ago
  3. lgbasallote Group Title
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    okay first turn dy/ sec^2 y into cos^2 y dy you agree those are the same right?

    • one year ago
  4. lgbasallote Group Title
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    \[\implies \int \cos^2 y dy = \int \frac{dx}{1+x^2}\]

    • one year ago
  5. haganmc Group Title
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    yes now i see.

    • one year ago
  6. lgbasallote Group Title
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    wonderful

    • one year ago
  7. haganmc Group Title
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    thanks

    • one year ago
  8. lgbasallote Group Title
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    welcome

    • one year ago
  9. haganmc Group Title
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    what about 1/(1+x^2)

    • one year ago
  10. lgbasallote Group Title
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    trigonometric substitution

    • one year ago
  11. haganmc Group Title
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    is it tan?

    • one year ago
  12. lgbasallote Group Title
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    arctan*

    • one year ago
  13. haganmc Group Title
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    right

    • one year ago
  14. lgbasallote Group Title
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    yes

    • one year ago
  15. haganmc Group Title
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    that's what I thought. Thanks

    • one year ago
  16. lgbasallote Group Title
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    welcome

    • one year ago
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