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Math Help - High oreder differential equation

  1. #1
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    High oreder differential equation

    Simple equation, but I need step by step analytical way. Thank you

    x''''[t] + r*x''[t] + v*x[t] = p[t]
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  2. #2
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    Quote Originally Posted by derdack View Post
    Simple equation, but I need step by step analytical way. Thank you

    x''''[t] + r*x''[t] + v*x[t] = p[t]
    You will need to solve the homogeneous and non-homogeneous solutions.

    Start by solving:

    x^{(4)}+rx''+vx=0\Rightarrow m^4+rm^2+v=0
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    Yes, but I know homogeneous solutions and way, but how to find non-homogeneous?
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    Quote Originally Posted by derdack View Post
    Yes, but I know homogeneous solutions and way, but how to find non-homogeneous?
    It depends on the form of p(t).

    If we assume

    p(t)=c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0

    then

    x(t)=A_nx^n+A_{n-1}x^{n-1}+\cdots+A_1x+A_0

    Now, solve and equate coefficients how x^{(4)}+rx^2+vx=p(t).

    This is one thought I have. Do you know anything about p(t)?
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  5. #5
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    We need to apply Lagrange method variational constants. p(t) time function. Math program give a solutions but how, step by step. It didn't need aproximation. I need analytical method.
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  6. #6
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    We can premise p(t) like a linear function of t
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  7. #7
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    However p(t) is unknown function!
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  8. #8
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    Quote Originally Posted by derdack View Post
    We can premise p(t) like a linear function of t
    If p(t) is linear it must be of the form p(t) = at + b. Can you do something with this?

    -Dan
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  9. #9
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    I am sure that a p is unknown function. I have homogenous solution. First derivate of x, constants have a derivate to (because of Lagrange method of variational constants). From this, velocity must be like a first derivate of homogenous solution when constants not have a derivate. We can get first equation from that like f(C1'(t), C2'(t))=0. Third derivate of hom solution can give us one more equation. Second and fourth derivate we can substitute in differential equation, but what to equal in this equation when we didn't know p. And we have 3 equation, and 4 unknown C1', C2',C3',C4'.
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  10. #10
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by derdack View Post
    Simple equation, but I need step by step analytical way. Thank you

    x''''[t] + r*x''[t] + v*x[t] = p[t]
    This type of problem is well trated with the Laplace Tranform. If x(t) and p(t) are L-tranformable and X(s) and P(s) are their L-tranforms we can write...

    \displaystyle \mathcal{L} \{x^{(2)} (t)\} = s^{2}\ X(s) -s\ x(0) - x^{'} (0)

    \displaystyle \mathcal{L} \{x^{(4)} (t)\} = s^{4}\ X(s) -s^{3}\ x(0) - s^{2}\ x^{'} (0) - s\ x^{(2)}(0) - x^{(3)} (0) (1)

    ... so trhat the DE in terms of L-tranform becomes...

    \displaystyle \mathcal{L} \{x^{(4)} (t)\} + r\ \mathcal{L} \{x^{(2)} (t)\} + v\ X(s) = P(s) (2)

    The (2) permits to find X(s) and the solution is x(t) = \mathcal{L}^{-1} \{X(s)\}...

    Kind regards

    \chi \sigma
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  11. #11
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    Quote Originally Posted by derdack View Post
    We need to apply Lagrange method variational constants. p(t) time function. Math program give a solutions but how, step by step. It didn't need aproximation. I need analytical method.
    So you still need to find the complimentary solution. Suppose that it has the form

    y_c=c_1y_1+c_2y_2+c_3y_3+c_4y_4

    Using variation of parameters we can use the wronskian to write a system to be solved.

    y_p=u_1y_1+u_2y_2+u_3y_3+u_4y_4

    \displaystyle  \begin{bmatrix} y_1 & y_2 & y_3 & y_4 \\ y_1' & y_2' & y_3' & y_4' \\ y_1'' & y_2'' & y_3'' & y_4'' \\ y_1''' & y_2''' & y_3''' & y_4'''  \end{bmatrix}\begin{bmatrix}u_1' \\ u_2' \\ u_3' \\ u_4' \end{bmatrix}=\begin{bmatrix}0 \\ 0 \\ 0 \\ p(t) \end{bmatrix}

    Now using Cramer's rule we can solve for for the derivative of each of the u's

    u_1'=\frac{\begin{vmatrix}0 & y_2 & y_3 & y_4 \\ 0 & y_2' & y_3' & y_4' \\ 0 & y_2'' & y_3'' & y_4'' \\ p(t) & y_2''' & y_3''' & y_4'''   \end{vmatrix}}{\begin{vmatrix} y_1 & y_2 & y_3 & y_4 \\ y_1' & y_2' & y_3' & y_4' \\ y_1'' & y_2'' & y_3'' & y_4'' \\ y_1''' & y_2''' & y_3''' & y_4'''  \end{vmatrix}}

    u_2'=\frac{\begin{vmatrix}y_1 & 0 & y_3 & y_4 \\ y_1' & 0 & y_3' & y_4' \\ y_1'' & 0 & y_3'' & y_4'' \\ y_1''' & p(t) & y_3''' & y_4'''   \end{vmatrix}}{\begin{vmatrix} y_1 & y_2 & y_3 & y_4 \\ y_1' & y_2' & y_3' & y_4' \\ y_1'' & y_2'' & y_3'' & y_4'' \\ y_1''' & y_2''' & y_3''' & y_4'''  \end{vmatrix}}

    and continue the pattern to find u_3',u_4'

    Now we can integrate (or write in terms of integrals) the particular solution.
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  12. #12
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    Thank you a lot... but i must find solutions in time regime, without Laplace t. If you know how. Anyway, thank you!!!
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  13. #13
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    Thank you a lot Mr Set. Thank you. You help me so much.
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  14. #14
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    "Undetermined Coefficients" works as long as p(t) is one of the kinds of functions we expect as a solution to a linear equation with constant coefficients- exponentials, sine and cosine, polynomials and products of those.

    For a general p(t), us "variation of parameters": If y_1(t) and y_2(t) are two independent solutions to the associated homogeneous equation, look for a solution to the entire equation of the form y(t)= u(t)y_1(t)+ v(t)y_2(t). The derivative of that is y'(t)= u'(t)y_1(t)+ u(t)y_1'(t)+ v'(t)y_2(t)+ v(t)y_2'(t). In fact, there are an infinite number of such solutions. We "narrow the search" but adding the requirement that the terms involving derivatives of u and v sum to 0: u'(t)y_1(t)+ v'(t)y_2(t)= 0.

    Now we have y'(x)= u(t)y_1'(t)+ v(t)y_2'(t) and, differentiating again, y''(t)= u'(t)y_1'(t)+ u(t)y_1''(t)+ v'(t)y_2'(t)+ v(t)y_2''(t). When we put those into the original differential equation, those terms involving u and v, rather than u' and v', will cancel. That is because, since u and v are not being differentiated, they are "like constants" and any constants times y_1(t) and y_2(t) satisfy the associated homogeneous equation.

    That means that we get an equation involving only known functions, p(t) and y_1(x)] and y_2(x) and their derivatives, with "unknowns", u' and v'. That, together with the equation u'y_1(x)+ v'y_2(x) gives two equations that can be solved algebraically for u' and v'.
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  15. #15
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    Thank you a lot! I will apply some of good answers!!! Great!!!
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