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$x(t)+\int_0^t(t-\tau)x(\tau)=t^2$

Is $x(\tau)$ the equivalent of $d\tau$? How do I solve this particular equation?

Lugi
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3 Answers3

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The $\,d\tau$ was just missing. That integral $\displaystyle\int_{0}^{t}(t-\tau)x(\tau)\,d\tau$ is $t$ convolved with $x(t)$.

Your first step should be to take the Laplace transform of both sides. Remember convolution in the time domain corresponds to multiplication in the Laplace domain.

$x(t) + \displaystyle\int_{0}^{t}(t-\tau)x(\tau)\,d\tau = t^2$

$X(s) + \dfrac{1}{s^2}X(s) = \dfrac{2}{s^3}$.

Now solve for $X(s)$.

JimmyK4542
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  • I solved it and got $2(cos t -1)$. Is that the good answer? Shouldnt there be any arbitrary constant in it? – Lugi Jun 29 '14 at 20:34
  • Try plugging in $x(t) = 2(\cos t - 1)$ and $x(t) = 2(\cos t - 1)+100$ into the integral equation and see which of them work. – JimmyK4542 Jun 29 '14 at 20:35
  • Actually neither of them. I missed a minus sign somewhere. – Lugi Jun 29 '14 at 20:43
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You can use Laplace transform techniques. See (I), (II).

Community
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Mhenni Benghorbal
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0

Hint:

Define: $$y(t):=x(t)+\int_0^t(t-\tau)x(\tau)-t^2=0......(1)$$ Then $$y''(t)=x''(t)+x(t)-2=0......(2)$$

You may then solve (2) for $x(t)$.

$$x(t)=2+c_1 \cos t+c_2\sin t$$

mike
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