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    The question is:

    Code:
    Find y in terms of x given that
    
    dy/dx + 2y = sinx
    
    and that the solution curve passes through the origin.
    I have got an answer which I think is right. But the answer in the back is different.
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    Maybe they're the same but rearranged... what have you got and what have they got?
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    (Original post by mikesgt2)
    The question is:

    Code:
    Find y in terms of x given that
    
    dy/dx + 2y = sinx
    
    and that the solution curve passes through the origin.
    I have got an answer which I think is right. But the answer in the back is different.
    I = e^(integral 2 dx)
    I = e^2x

    e^2x dy/dx + 2e^2x y = e^2x sinx

    e^2x y = integral e^2x sinx dx == K

    u=e^2x => du/dx = 2e^2x
    dv/dx = sinx => v=-cosx

    K = -e^2x cosx + 2 {integral e^2x cosx dx == L}

    a = e^2x => da/dx= 2 e^2x
    db/dx = cosx => b=sinx

    L = e^2x sinx - 2 integral e^2x sinx == e^2x sinx - 2K

    K = -e^2x cosx + 2 e^2x sinx - 4K
    5K = 2 e^2x sinx - e^2x cosx = e^2x (2six - cosx)

    => e^2x y = 1/5 e^2x (2sinx - cosx)

    => y = 1/5 (2sinx - 1osx + C)

    0 = 2/5 sin 0 -1/5 sin 0 + 1/5 C

    => C = 1

    => y = 1/5 (2sinx - cosx +1)
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    (Original post by mikesgt2)
    The question is:

    Code:
    Find y in terms of x given that
    
    dy/dx + 2y = sinx
    
    and that the solution curve passes through the origin.
    I have got an answer which I think is right. But the answer in the back is different.
    It's an integrating factor question with your IF being e^2x isn't it?
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    Elpaw...

    I got what you got... apart from I just realised I made a stupid mistake in calculating the constant of integration.

    But... the answer in the back is:

    y = 1/5 ( 2sinx - cosx +e^(-2x) )
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    (Original post by mikesgt2)
    Elpaw...

    I got what you got... apart from I just realised I made a stupid mistake in calculating the constant of integration.

    But... the answer in the back is:

    y = 1/5 ( 2sinx - cosx +e^(-2x) )
    yeah i se where that came from:

    => e^2x y = 1/5 e^2x (2sinx - cosx) + C

    => y = 1/5 (2sinx - cosx) + C e^-2x

    dot dot dot
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    Oh rite, I see. Thanks for your help.
 
 
 
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