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Math Help - 2 more questions

  1. #1
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    2 more questions

    once the integral is found for dy/dx=y^2+1, how do you find the equation that satisfies the initial condition of y(4)=0?

    i put y=tan(x+4), but it's wrong.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by sw3etazngyrl View Post
    once the integral is found for dy/dx=y^2+1, how do you find the equation that satisfies the initial condition of y(4)=0?

    i put y=tan(x+4), but it's wrong.
    i got y = tan(x - 4). what was the answer?
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  3. #3
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    i don't know, it doesn't tell me. it only tells me that my answer is wrong.
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  4. #4
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    how did you get that answer?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by sw3etazngyrl View Post
    i don't know, it doesn't tell me. it only tells me that my answer is wrong.
    oh. you're inputing answers into some online application? is y = tan(x - 4) also wrong?

    we have y' = y^2 + 1

    \Rightarrow \frac {y'}{y^2 + 1} = 1

    \Rightarrow \arctan y = x + C

    \Rightarrow y = \tan(x + C)

    Now y(4) = 0

    \Rightarrow 0 = \tan (4 + C)

    \Rightarrow \arctan0 = 4 + C

    \Rightarrow 0 = 4 + C

    \Rightarrow C = -4

    Thus our solution is y = \tan (x - 4)
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  6. #6
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    the answer you got is right. thanks! but how did you arrive at that?
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  7. #7
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    \frac{{dy}}<br />
{{dx}} = y^2 + 1 \implies \frac{1}<br />
{{y^2 + 1}}\,dy = dx \implies \arctan y = x + k

    It's just arctan formula.
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