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Thread: Inhomogenous differential Equations

  1. #1
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    Inhomogenous differential Equations

    Hello,

    I'm stumbling on an example for finding the particular solution to a difference equation. I have:

    $\displaystyle y'(t) - 3y(t) = t^2u(t) : y(0) = 2$

    So I understand that one way to solve this is by finding the homogenous solution in terms of unknown integration constants, then finding the particular solution. I've figured out the homogenous solution to be

    $\displaystyle Ae^tu(t)$

    But I'm tripping on the particular solution. Stroud's book 'Advanced Engineering Mathematics' states that if the right hand side of the original difference equation has a polynomial term $\displaystyle t^n$ then the particular solution should be in the form

    $\displaystyle C_{n+1}t^{n+1} + C_nt^n + C_{n-1}t^{n-1} + ... + C_0$

    So I take this that the particular solution should be of the form

    $\displaystyle Ct^3 + Dt^2 + Et + F$

    However in Stroud's example for $\displaystyle n^2$ he gives

    $\displaystyle Cn^2 + Dn + E$

    Have I misunderstood or has he made a mistake? Thanks!
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  2. #2
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    Re: Inhomogenous differential Equations

    Quote Originally Posted by halfnormalled View Post
    Hello,

    I'm stumbling on an example for finding the particular solution to a difference equation. I have:

    $\displaystyle y'(t) - 3y(t) = t^2u(t) : y(0) = 2$

    So I understand that one way to solve this is by finding the homogenous solution in terms of unknown integration constants, then finding the particular solution. I've figured out the homogenous solution to be

    $\displaystyle Ae^tu(t)$

    But I'm tripping on the particular solution. Stroud's book 'Advanced Engineering Mathematics' states that if the right hand side of the original difference equation has a polynomial term $\displaystyle t^n$ then the particular solution should be in the form

    $\displaystyle C_{n+1}t^{n+1} + C_nt^n + C_{n-1}t^{n-1} + ... + C_0$

    So I take this that the particular solution should be of the form

    $\displaystyle Ct^3 + Dt^2 + Et + F$

    However in Stroud's example for $\displaystyle n^2$ he gives

    $\displaystyle Cn^2 + Dn + E$

    Have I misunderstood or has he made a mistake? Thanks!
    I seriously doubt that $\displaystyle \displaystyle y = Ae^tu(t)$ could possibly be a solution. Have you tried substituting this into the DE and seeing if the equation is balanced?

    You can solve this using the Integrating Factor method, since it's first order linear.

    $\displaystyle \displaystyle \frac{dy}{dt} - 3y = t^2u(t)$

    The integrating factor is $\displaystyle \displaystyle e^{\int{-3\,dt}} = e^{-3t}$, so multiplying both sides gives

    $\displaystyle \displaystyle \begin{align*} e^{-3t}\frac{dy}{dt} - 3e^{-3t}y &= t^2e^{-3t}u(t) \\ \frac{d}{dt}\left(e^{-3t}y\right) &= t^2e^{-3t}u(t) \\ e^{-3t}y &= \int{t^2e^{-3t}u(t)\,dt} + C \\ y &= e^{3t}\int{t^2e^{-3t}u(t)\,dt} + Ce^{3t} \end{align*}$
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  3. #3
    MHF Contributor chisigma's Avatar
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    Re: Inhomogenous differential Equations

    Quote Originally Posted by halfnormalled View Post
    Hello,

    I'm stumbling on an example for finding the particular solution to a difference equation. I have:

    $\displaystyle y'(t) - 3y(t) = t^2u(t) : y(0) = 2$


    So I understand that one way to solve this is by finding the homogenous solution in terms of unknown integration constants, then finding the particular solution. I've figured out the homogenous solution to be

    $\displaystyle Ae^tu(t)$

    But I'm tripping on the particular solution. Stroud's book 'Advanced Engineering Mathematics' states that if the right hand side of the original difference equation has a polynomial term $\displaystyle t^n$ then the particular solution should be in the form

    $\displaystyle C_{n+1}t^{n+1} + C_nt^n + C_{n-1}t^{n-1} + ... + C_0$

    So I take this that the particular solution should be of the form

    $\displaystyle Ct^3 + Dt^2 + Et + F$

    However in Stroud's example for $\displaystyle n^2$ he gives

    $\displaystyle Cn^2 + Dn + E$

    Have I misunderstood or has he made a mistake? Thanks!
    Writing the DE in terms of Laplace Tranform You obtain...

    $\displaystyle s\ Y(s) - y(0) -3\ Y(s) = \frac{2}{s^{3}}$ (1)

    ... and from (1) You derive the explicit expression of Y(s)...

    $\displaystyle Y(s)= \frac{y(0)}{s-3} + \frac{2}{s^{3}\ (s-3)}$ (2)

    The second fraction can be expanded as...

    $\displaystyle \frac{2}{s^{3}\ (s-3)}= - \frac{\frac{2}{27}}{s} - \frac{\frac{2}{9}}{s^{2}} - \frac{\frac{2}{3}}{s^{3}} + \frac{\frac{2}{27}}{s-3}$ (3)

    ... so that the solution is...

    $\displaystyle y(t)= (\frac{56}{27}\ e^{3 t} - \frac{2}{27} - \frac{2}{9}\ t - \frac{1}{3}\ t^{2})\ \mathcal{U} (t)$ (4)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  4. #4
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    Re: Inhomogenous differential Equations

    Thank you for your replies, and sorry mods for putting this in the wrong forum

    I understand the workings you provided, but I'm not sure they are the right application for my problem. My fault for not asking the question correctly. The question is to do with linear systems and finding the zero state response, then the zero state response. Then to find the complete solution which I understand is:

    $\displaystyle y(t) = y_{zs}(t) + y_{zi}(t)$

    I gather that I should find the homogenous (or complementary as it seems to be referred to in some places), then find the particular solution. so that the complete solution is

    $\displaystyle y(t) = y_h(t) + y_p(t)$

    Then, I think I understand that in order to find the zero input response I need to apply the original boundary conditions to $\displaystyle y_h(t)$. This is why I think I need to find the homogenous or complementary solution first. Then I need to find the particular solution. Correct?

    So here is my working for the particular solution: (which by the time I'd got to the end of this post, I think I've figured out )

    Based on the form of the RHS of the original equation it should take the form
    $\displaystyle Ct^2 + Dt + e$ (which is what my original query was supposed to be about.. is this right?)

    $\displaystyle \frac{d}{dt}Ct^2 + Dt + e = 2Ct + D$

    so substituting in the original equation:

    $\displaystyle 2Ct + D - 3Ct^2 - 3Dt - 3e = t^2$

    gives

    $\displaystyle C = -\frac{1}{3}, D = -\frac{2}{9}, e = -\frac{2}{27}$

    so

    $\displaystyle y_p(t) = -\frac{1}{3}t^2 - \frac{2}{9}t - \frac{2}{27}$
    $\displaystyle = -\frac{1}{27}(9t^2 + 6t + 2)$

    $\displaystyle y(t) = y_h(t) + y_p(t)$

    $\displaystyle y(t) = Ae^{3t} - \frac{1}{27}(9t^2 + 6t + 2)$

    Then the zero state response is found by setting boundary condition to zero i.e. $\displaystyle f(0) = 0$

    $\displaystyle 0 = Ae^{(3)(0)} - \frac{1}{27}(9(0)^2 + 6(0) + 2)$
    $\displaystyle 0 = A - \frac{1}{27}(2)$

    $\displaystyle y_{zs}(t) = \frac{2e^{3t}}{27} - \frac{1}{27}(9t^2 + 6t + 2)$
    $\displaystyle y_{zs}(t) = \frac{1}{27}\left\{2e^{3t} - (9t^2 + 6t + 2)\right\}$

    And the zero input response is achieved by setting the original boundary conditions in $\displaystyle y_h(t)$, so:

    $\displaystyle y_h(t) = Ae^{3t}$

    apply conditions provided as $\displaystyle y(0) = 2$

    $\displaystyle 2 = Ae^{3(0)}$

    $\displaystyle A = 2$

    $\displaystyle y_{zi}(t) = 2e^{3t}$

    So summing the zero state and zero input should give the complete solution:

    $\displaystyle Y(t) = 2e^{3t} + \frac{1}{27}\left\{2e^{3t} - (9t^2 + 6t + 2)\right\}$

    which matches up with chisigma's solution:

    $\displaystyle Y(t) = \frac{1}{27}\left\{56e^{3t} - (9t^2 + 6t + 2)\right\}$

    phew... I thought I'd put this up in case it helps someone else. And it has confirmed my original confusion about how the particular solution form was found. Thanks guys for your input.

    --------

    p.s.

    Oh, and yes, my original post had a mistake in. The homogenous solution $\displaystyle y_h(t)$ should have been $\displaystyle Ae^{3t}u(t)$. the $\displaystyle u(t)$ is to ensure that $\displaystyle y_h(t) = 0 for t<0$. I've put this back in and validated it (I think)...

    $\displaystyle y'(t) - 3y(t) = 0$
    $\displaystyle (Ae^{3t})' - 3Ae^{3t} = 0$
    $\displaystyle 3Ae^{3t} - 3Ae^{3t} = 0$
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