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Differential equations assignment T5

  1. Dec 4, 2013 #1
    Hi!

    I would like to ask anyone with some spare time to check my assignment questions. Last time I was asked to post one task at a time so I will.
    Thank you in advance for your time.

    Task 5:

    Find the particular solution of the following differential equations:
    a) 12(d2y/dx2)-3y=0
    given that: x=0, y=3 and (dy/dx)=0.5

    b) (d2y/dx2)+2(dy/dx)+2y=10ex
    given that: x=0, y=0 and (dy/dx)=1

    Solution:
    a)
    12(d2y/dx2)-3y=0 /:12
    (d2y/dx2)-(1/4)y=0

    m2=n2
    m=+/-n

    ∴y=Aenx+Be-nx
    or y=Acosh(nx)+Bsinh(nx)

    m2=1/4
    ∴m=+/-√(1/4)=+/-(1/2)

    y=Acosh[(1/2)x]+Bsinh[(1/2)x]

    3=Acosh[0]+Bsinh[0]
    3=A(e0+e0/2)+B(e0-e0/2)
    A=3

    dy/dx=Asinh(0)+Bcosh(0)
    B=1/2

    Solution
    b)
    d2y/dx2+2(dy/dx)+2y=10ex

    m2+2m+2=0
    ∴m=-1+/-j

    CF:
    u=e-x{Acos(x)+Bsin(x)}

    PI:
    v=pex
    v'=pex
    v''=pex

    pex+2pex+2pex=10ex
    5pex=10ex /:5ex
    p=10ex/5ex=2

    v=2ex

    GS:
    y=e-x{Acos(x)+Bsin(x)}+2ex [1]

    dy/dx=e-x{-Asin(x)+Bcos(x)}-e-x{Acos(x)+Bsin(x)}+2ex [2]

    Sub into [1]
    0=e0{Acos(0)+Bsin(0)}+2e0
    A=-2

    Sub into [2]
    1=e0{2sin(0)+Bcos(0)}-e0{2cos(0)+Bsin(0)}+2e0
    1=B-2+2
    B=1

    PS:
    y=e-x{-2cos(x)+sin(x)}+2ex
     
    Last edited: Dec 4, 2013
  2. jcsd
  3. Dec 4, 2013 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    This is correct but I don't see why you would change to exponential form. If you are using hyperbolic functions you should know that a good reason for doing that is that cosh(0)= 1 and sinh(0)= 0. 3= Acosh(0)+ Bsinh(0) is immediately "3= A".

    Yes, dy/dx(0)= B= 0.5
    Now, how about actually writing the solution to the problem?

    You've lost a sign here: the second term should be -e0(-2cos(0)+ Bsin(0))

    So this should be 1= B+ 2+ 2

     
  4. Dec 4, 2013 #3
    Thank you for the reply!

    I'm actually not sure why I did it.

    y=3cosh(x/2)+0.5sinh(x/2)

    B=-3

    ∴y=e-x{-2cos(x)-3sin(x)}+2ex
     
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