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Thread: System of DE's.

  1. #1
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    System of DE's.

    Please help. Solve
    dx/dt + y = sint , x(0)=2
    dy/dt + x = cost , y(0)=0

    this is one problem that comes in my laplace transform chapter. Any method can be used
    Last edited by mr fantastic; Nov 26th 2010 at 09:18 PM. Reason: Re-titled.
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  2. #2
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    From DE 1: $\displaystyle \displaystyle \frac{dx}{dt} + y = \sin{t}$

    $\displaystyle \displaystyle y = -\frac{dx}{dt} + \sin{t}$

    $\displaystyle \displaystyle \frac{dy}{dt} = -\frac{d^2x}{dt^2} + \cos{t}$.


    Substituting into DE 2:

    $\displaystyle \displaystyle -\frac{d^2x}{dt^2} + \cos{t} + x = \cos{t}$

    $\displaystyle \displaystyle -\frac{d^2x}{dt^2} + x = 0$

    $\displaystyle \displaystyle \frac{d^2x}{dt^2} - x = 0$.


    Now, the characteristic equation is $\displaystyle \displaystyle m^2 - 1 = 0$

    $\displaystyle \displaystyle m^2 = 1$

    $\displaystyle \displaystyle m = \pm 1$.


    So the solution to the homogeneous equation is $\displaystyle \displaystyle x = Ae^{t} + Be^{-t}$.

    Follow a similar process to find $\displaystyle \displaystyle y$.
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    Quote Originally Posted by Prove It View Post
    From DE 1: $\displaystyle \displaystyle \frac{dx}{dt} + y = \sin{t}$

    $\displaystyle \displaystyle y = -\frac{dx}{dt} + \sin{t}$

    $\displaystyle \displaystyle \frac{dy}{dt} = -\frac{d^2x}{dt^2} + \cos{t}$.


    Substituting into DE 2:

    $\displaystyle \displaystyle -\frac{d^2x}{dt^2} + \cos{t} + x = \cos{t}$

    $\displaystyle \displaystyle -\frac{d^2x}{dt^2} + x = 0$

    $\displaystyle \displaystyle \frac{d^2x}{dt^2} - x = 0$.


    Now, the characteristic equation is $\displaystyle \displaystyle m^2 - 1 = 0$

    $\displaystyle \displaystyle m^2 = 1$

    $\displaystyle \displaystyle m = \pm 1$.


    So the solution to the homogeneous equation is $\displaystyle \displaystyle x = Ae^{t} + Be^{-t}$.

    Follow a similar process to find $\displaystyle \displaystyle y$.
    But if we do like this , we may not be able to make use of the condition given in the question. Can i ?. Please respond. Thank you
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    Quote Originally Posted by rafi98 View Post
    But if we do like this , we may not be able to make use of the condition given in the question. Can i ?. Please respond. Thank you
    Find $\displaystyle \displaystyle y$ first, see what you get.
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    What is small letter a and b . And i dont know the formula of your second reply.
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    Quote Originally Posted by rafi98 View Post
    Please help. Solve
    dx/dt + y = sint , x(0)=2
    dy/dt + x = cost , y(0)=0

    this is one problem that comes in my laplace transform chapter. Any method can be used
    From the second DE

    $\displaystyle \displaystyle \frac{dy}{dt} + x = \cos{t}$

    $\displaystyle \displaystyle x = -\frac{dy}{dt} + \cos{t}$

    $\displaystyle \displaystyle \frac{dx}{dt} = -\frac{d^2y}{dt^2} - \sin{t}$.

    Substitute into the first DE and solve for $\displaystyle \displaystyle y$.
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  7. #7
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    Quote Originally Posted by rafi98 View Post
    Please help. Solve
    dx/dt + y = sint , x(0)=2
    dy/dt + x = cost , y(0)=0

    this is one problem that comes in my laplace transform chapter. Any method can be used
    apply the Laplace Transform and get

    $\displaystyle s\widehat x(s)-x(0)+\widehat y(s)=\dfrac1{s^2+1}$ and $\displaystyle s\widehat y(s)-y(0)+\widehat x(s)=\dfrac s{s^2+1},$ the rest is algebra and a matter of finding inverse transforms.
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