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haganmc

  • 3 years ago

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

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  1. lgbasallote
    • 3 years ago
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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?

  2. haganmc
    • 3 years ago
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    yes. the Integration is the problem i'm having trouble with

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

  4. lgbasallote
    • 3 years ago
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    \[\implies \int \cos^2 y dy = \int \frac{dx}{1+x^2}\]

  5. haganmc
    • 3 years ago
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    yes now i see.

  6. lgbasallote
    • 3 years ago
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    wonderful

  7. haganmc
    • 3 years ago
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    thanks

  8. lgbasallote
    • 3 years ago
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    welcome

  9. haganmc
    • 3 years ago
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    what about 1/(1+x^2)

  10. lgbasallote
    • 3 years ago
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    trigonometric substitution

  11. haganmc
    • 3 years ago
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    is it tan?

  12. lgbasallote
    • 3 years ago
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    arctan*

  13. haganmc
    • 3 years ago
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    right

  14. lgbasallote
    • 3 years ago
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    yes

  15. haganmc
    • 3 years ago
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    that's what I thought. Thanks

  16. lgbasallote
    • 3 years ago
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    welcome

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