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Math Help - Differential Equation help

  1. #1
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    Differential Equation help

    Hi im a bit stuck on this question any help would be appreciated

    solve the differential equation dy/dx + ycosx/sin x = x, subject to the initial condition y(pi/2) = 3
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  2. #2
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    First, find your integrating factor.

    cos(x)/sin(x)=cot(x)

    e^(\int cot(x)dx)=sin(x)

    d/dx[ysin(x)]=xsin(x)

    Integrate:

    ysin(x)=sin(x)-xcos(x)+C

    Divide by sin(x):

    y=1-xcot(x)+Ccsc(x)

    Now, use your IC to find C and you're done.
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  3. #3
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    Wink

    given

    dy/dx+(cosx/sinx)y=x

    dy/dx+py=q
    which is linear differential equation of first order
    here p=cosx/sinx and q=x

    Integrating factor= e^int(cosx/sinx)=e^log|sinx| = sinx

    solution of the given equation is
    y(sinx)=xsinx+c
    since y(pi/2)=3

    3(sin(pi/2))=pi/2sin(pi/2)+c
    3=pi/2+c
    3-pi/2=c

    hence the solution is
    y(sinx)=xsinx+3-pi/2 is the answer
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  4. #4
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    answer

    Hi,
    given

    dy/dx+(cosx/sinx)y=x

    dy/dx+py=q
    which is linear differential equation of first order
    here p=cosx/sinx and q=x

    Integrating factor= e^int(cosx/sinx)=e^log|sinx| = sinx

    solution of the given equation is
    y(sinx)=xsinx+c
    since y(pi/2)=3

    3(sin(pi/2))=pi/2sin(pi/2)+c
    3=pi/2+c
    3-pi/2=c

    hence the solution is
    y(sinx)=xsinx+3-pi/2 is the answer __________________
    RK
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