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Math Help - Solving a differential equation using Laplace transform

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    Solving a differential equation using Laplace transform

    Hi there. I have to solve the differential equation using the laplace transform.

    ty''+2y'+ty=0,y(0)=1,y(\pi)=0

    So, applying the Laplace transform on both sides of the equality, and using some properties of the Laplace transform I get:

    -\frac{d}{ds}(s^2y(s)-s-y'(0))+2sy(s)-2y(0)-\frac{dy}{ds}=0

    -s^2\frac{dy}{ds}-\frac{dy}{ds}-1=0\rightarrow -\frac{ds}{s^2+1}=dy

    Then: y(s)=-\arctan(s)+C

    Then I have to apply the inverse transform to get the function I'm looking for, but the thing is I don't know the inverse Laplace transform for the arctangent of s. I think I have probably done something wrong, but I couldn't find the mistake.
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    Re: Solving a differential equation using Laplace transform

    Quote Originally Posted by Ulysses View Post
    Hi there. I have to solve the differential equation using the laplace transform.

    ty''+2y'+ty=0,y(0)=1,y(\pi)=0

    So, applying the Laplace transform on both sides of the equality, and using some properties of the Laplace transform I get:

    -\frac{d}{ds}(s^2y(s)-s-y'(0))+2sy(s)-2y(0)-\frac{dy}{ds}=0

    -s^2\frac{dy}{ds}-\frac{dy}{ds}-1=0\rightarrow -\frac{ds}{s^2+1}=dy

    Then: y(s)=-\arctan(s)+C

    Then I have to apply the inverse transform to get the function I'm looking for, but the thing is I don't know the inverse Laplace transform for the arctangent of s. I think I have probably done something wrong, but I couldn't find the mistake.
    Yes you have definitely done something wrong. When you take the Laplace Transform, you should get some function of s, not a differential equation that involves s.
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    Re: Solving a differential equation using Laplace transform

    Thats because the differential equation has variable coefficients, so one gets a differential equation of s. I've seen other exercises likes this. Perhaps I wasn't clear, the function I'm looking is y(t), y(s) denotes it's Laplace transform.

    When I apply the laplace transform on the equation I get:
    \mathcal{L} \left [ ty'' \right ] +2 \mathcal{L} \left [ y' \right ]+ \mathcal{L} \left [ ty \right ]=0

    Then using \mathcal{L} \left [t^n F(t)\right ]=(-1)^n\frac{d^n f}{ds^n}(s)
    And \mathcal{L} \left [ F^n(t) \right ]=s^n \mathcal{L} \left [ F(t)\right ]-s^{n-1}F(0)-s^{n-2}F'(0)-...-F^{n-1}(0)

    I get for example:
    \mathcal{L} \left [ ty'' \right ]=-\frac{d}{ds}(s^2 \mathcal{L}\left [ y \right ]-sy(0)-y'(0))
    I don't know y'(0), but its a constant, then when deriving I get to the second equation that I put at the beginning.
    Last edited by Ulysses; November 22nd 2011 at 04:17 PM.
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    Re: Solving a differential equation using Laplace transform

    Quote Originally Posted by Ulysses View Post
    Hi there. I have to solve the differential equation using the laplace transform.

    ty''+2y'+ty=0,y(0)=1,y(\pi)=0

    So, applying the Laplace transform on both sides of the equality, and using some properties of the Laplace transform I get:

    -\frac{d}{ds}(s^2y(s)-s-y'(0))+2sy(s)-2y(0)-\frac{dy}{ds}=0

    -s^2\frac{dy}{ds}-\frac{dy}{ds}-1=0\rightarrow -\frac{ds}{s^2+1}=dy

    Then: y(s)=-\arctan(s)+C

    Then I have to apply the inverse transform to get the function I'm looking for, but the thing is I don't know the inverse Laplace transform for the arctangent of s. I think I have probably done something wrong, but I couldn't find the mistake.
    You got \frac{dF}{ds} = \frac{-1}{s^2 + 1} which is correct.

    Now note that LT[t f(t)] = - \frac{dF}{ds} and so you have LT[t f(t)] = \frac{1}{s^2 + 1} \Rightarrow t f(t) = ....
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