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Math Help - Specific Solution for a Differential Equation

  1. #1
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    Specific Solution for a Differential Equation

    Find the solution of the diff eq that satisfies the given initial condition.

    y'tan x = a + y, y(pi/4) = a, 0 < x < pi/2

    y' = (a+y)/tanx

    dy/ (a+y) = cot x dx

    Integrate...

    ln (a + y ) = ln (sinx ) + c

    I think to make this work, c will be 2a- root2/2, so my answer is
    looks like

    y = sinx - a + 2a - root2 /2

    Is this correct?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by veronicak5678 View Post
    Find the solution of the diff eq that satisfies the given initial condition.

    y'tan x = a + y, y(pi/4) = a, 0 < x < pi/2

    y' = (a+y)/tanx

    dy/ (a+y) = cot x dx

    Integrate...

    ln (a + y ) = ln (sinx ) + c

    I think to make this work, c will be 2a- root2/2, so my answer is
    looks like

    y = sinx - a + 2a - root2 /2

    Is this correct?

    When solving for y, you should end up with a+y=e^{\ln(\sin x)+C}\implies y=C\sin x-a

    When you apply the initial condition, you should get C=2\sqrt{2}a

    Thus, \color{red}\boxed{y=2\sqrt{2}a\sin x-a}

    Does this make sense?

    --Chris
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  3. #3
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    No... could you show me how to solve for y?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by veronicak5678 View Post
    No... could you show me how to solve for y?
    Your work is good up to here:

    \ln(a+y)=\ln(\sin x)+C

    To get rid of the logarithm on the left side of the equation, apply the exponential function:

    \ln(a+y)=\ln(\sin x)+C\implies e^{\ln(a+y)}=e^{\ln(\sin x)+C}\implies a+y=e^{\ln(\sin x)+C}

    Thus,

    a+y=e^{\ln(\sin x)+C}\implies a+y=e^{\ln(\sin x)}e^C\implies a+y=Ce^{\ln(\sin x)}; e^C=C

    Finally,

    a+y=Ce^{\ln(\sin x)}\implies a+y=C\sin x\implies y=C\sin x-a

    Now plug in the initial condition to find C.

    Does this clarify things?

    --Chris
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  5. #5
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    Got it. I would never have known to do that! Thanks a lot.
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