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anonymous

  • one year ago

Verify that y=tan(x+c) is a one parameter family of solutions of the differential equation y'=1+y^2

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  1. anonymous
    • one year ago
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    y=tan(x+c) y'=sec^2(x+c)

  2. anonymous
    • one year ago
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    Answer: No real solutions.

  3. IrishBoy123
    • one year ago
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    keep going @Mateaus

  4. anonymous
    • one year ago
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    y'=sec^2(x+c) y'=tan^2(x+c)+1 that's how far I can simplify it but I am unsure if there is a possible way to get y'=1+y^2

  5. IrishBoy123
    • one year ago
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    ?? you have done it, you were given : y'=1+y^2 with: y=tan(x+c) you showed that : y'=sec^2(x+c) [= y^2 +1]

  6. anonymous
    • one year ago
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    oh! derp... I see. Did the same mistake for another problem lol

  7. anonymous
    • one year ago
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    Thanks!

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