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Math Help - Need urgent help on laplace transform and PDE !

  1. #1
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    Exclamation Need urgent help on laplace transform and PDE !

    I'm stuck with this 2 questions ...

    Q1) Need urgent help on laplace transform and PDE !-laplace-transform.jpg

    this is what i get after rewriting for the step function: 3sint [1-u(t-pi)] + (-3sint)u(t-pi) ... im lost from then on.

    Q2) Need urgent help on laplace transform and PDE !-pde.jpg

    i'm really at my wits end ... need to submit shortly afterwards. thanks a lot for the help.
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  2. #2
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    Here is 2a.
    Attached Thumbnails Attached Thumbnails Need urgent help on laplace transform and PDE !-picture9.gif  
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  3. #3
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    2b
    Attached Thumbnails Attached Thumbnails Need urgent help on laplace transform and PDE !-picture10.gif  
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  4. #4
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    Quote Originally Posted by reverie414 View Post
    I'm stuck with this 2 questions ...

    Q1) Click image for larger version. 

Name:	laplace transform.jpg 
Views:	48 
Size:	17.3 KB 
ID:	2530

    this is what i get after rewriting for the step function: 3sint [1-u(t-pi)] + (-3sint)u(t-pi) ... im lost from then on.

    Q2) Click image for larger version. 

Name:	PDE.jpg 
Views:	51 
Size:	24.7 KB 
ID:	2531

    i'm really at my wits end ... need to submit shortly afterwards. thanks a lot for the help.
    I learn how to do Q1 today.
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  5. #5
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    You expressed it correctly as a Heaviside function now you find the Laplacian of both sides.

    The right hand side becomes,

    [-y'(0)-sy(0)+s^2F(s)] + 4F(s) = WHATEVER.

    Where, "WHATEVER" is the Laplacian of: 3sin(t) [1-u(t-pi)] -3sin(t)u(t-pi)

    First you can write,
    3sin(t) - 6sin(t)*u(t-pi)

    Use trigonometric identity: sin(t) = -sin(t-pi)

    6sin(t-pi)*u(t-pi) + 3sin(t)

    Now the Laplace Transform of,
    L{u(t-c)*f(t-c)} = exp(-cs) * F(s)

    Thus,
    L{6sin(t-pi)*u(t-pi)} = 6*exp(-pi*s)*L{sin(t)}

    6*exp(-pi*s)*(1/(s^2+1^2))

    So now, you have found equation in terms of Laplace transforms you just solve for F(s) an find the inverse Laplace transform.
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  6. #6
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    thanks. may i know how do you derive this:

    3sin(t) - 6sin(t)*u(t-pi)
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  7. #7
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    Quote Originally Posted by reverie414 View Post
    thanks. may i know how do you derive this:

    3sin(t) - 6sin(t)*u(t-pi)
    You found this,
    3sin(t) [1-u(t-pi)] -3sin(t)u(t-pi)
    For the expression of the non-homogenous term.

    If you open paranthesis you get,
    3sin(t) - 3sin(t)*u(t-pi) - 3sin(t)u(t-pi)
    Now I combine them,
    3sin(t)-6sin(t)*u(t-pi)
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  8. #8
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    oic. i get what u mean. thanks.
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