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Math Help - d2y/dx2?

  1. #1
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    Post d2y/dx2?

    d2y/dx2=sec 2x, y(0)=0 and y'(0)=1

    so the first deritive is tanx+1
    because tan0+c=1
    c=1

    for the orinal equation does it go back to sec 2x+c?

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  2. #2
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     y = \int \tan(x)+1 ~dx= \int \frac{\sin(x)}{\cos(x)}+1 ~dx

    Can you take it from here?
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  3. #3
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    i know that the antiderivative of sinx/cosx is -ln cosx +c but wher does the +1 come in?
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  4. #4
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    Quote Originally Posted by genlovesmusic09 View Post

    so the first deritive is tanx+1
     y = \int \tan(x)+1 ~dx= \int \frac{\sin(x)}{\cos(x)}+1 ~dx= \int \frac{\sin(x)}{\cos(x)} ~dx+ \int 1 ~dx <br />

    with

     <br />
\int 1 ~dx = x+c<br />


     y = \int \frac{\sin(x)}{\cos(x)} ~dx+ \int 1 ~dx = -\ln(-\cos(x)) +x+c<br />


    Now use initial condition of y(0)=0 to solve for c.
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  5. #5
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    thanks! i forgot you can separate!
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