The equation

$\dfrac{Y(s)}{s^2} + \dfrac{Y'(s)}{s} = \dfrac{-a}{s^4}$

is in the Laplace transform. How can I take the inverse i.e transform back to time domain and solve for a?


2 Answers 2


$$s Y'(s) +Y(s) = \frac{d}{ds} (s Y(s)) = -\frac{a}{s^2} \implies sY(s) = \frac{a}{s} + C$$


$$Y(s) = \frac{a}{s^2} + \frac{C}{s} $$

Inverting the LT, we have

$$y(t) = a t +C$$

Information on $y(0)$ would determine $C$.


I just joined so I can't add a comment to Ron's answer but wanted to clarify that information on y for any t can be used to find C, not just y(0).


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