I have been given the following initial value problem.

$y'' + 4y = f(t)$


I need to use the Laplace transform to show that the answer is of the form:

$y=A\cos(2t) + B\sin(2t) + \frac{1}{2}\int_{0}^{t}f(\tau)\sin(2(t-\tau))d\tau$

I then need to deduce the value of the integral :


My working so far is using the normal method with the transform of both sides.

$s^2Y(s) - 3s + 1 + Y(s) = \int_{-infty}^{infty}e^{-st}f(t)dt$

Re-arranging we get:

$Y(s) = \frac{\int_{-\infty}^{\infty}e^{-st}f(t)dt}{s^2+1} + \frac{3s}{s^2 + 1} - \frac{1}{s^2 + 1}$

I get that the inverse transforms of the two fractions on the right-hand side are $3\cos(t)$ and $-\sin(t)$ but I am unsure what I am supposed to do with the integral of the arbitrary function or how to obtain the given result, as the sinusoidal terms look wrong anyway

Is there some result I am missing? I can see a shift in the final exponential so is the shift theorem involved somewhere? The textbook I am using is of no help and goes off on a tangent about circuits. The videos and examples from textbooks are just simple examples with 0 or an exponential on the RHS where you solve by doing a partial fraction decomposition to get the final answer.


1 Answer 1


Firstly : you have forgotten coefficient $4$ in your fifth equation which should be :

$$\tag{1}s^2Y(s) - 3s + 1 + 4Y(s) = F(s)$$

Comment: Instead of coming back to the Laplace Transform definition, I write as $F(s)$ the Laplace Transform of $f(t).$


$$\tag{2}Y(s) =3\frac{s}{s^2 + 2^2} - \dfrac{3}{2}\frac{2}{s^2 + 2^2}+ \dfrac{1}{2}F(s) \times \frac{2}{s^2+2^2}$$

Now, applying the inverse Laplace Transform to (2) we get:

$$y(t)=3\cos(2t) - \dfrac{3}{2}\sin(2t) + \frac{1}{2}f(t) \star \sin(2t) $$

because the inverse Laplace Transform of a product of two $s$-expressions is the convolution (notation $\star$) of their inverse Laplace Transforms.

Indeed, this convolution can be written :

$$f(t) \star \sin(2t) \ = \ \int_{0}^{t}f(\tau)\sin(2(t-\tau))d\tau,$$

the defining formula for the so-called one-sided convolution being :

$$f(t) \star g(t) \ = \ \int_{0}^{t}f(\tau)g(t-\tau)d\tau$$

See this MIT document.

Following our exchanges, here is a MatLab program displaying a numerical way to get a convolution; I have taken your example with $f(t):=t$ (it is true that in this case there is an explicit expression which is $\tfrac12t-\tfrac14 \sin(2t)$).

 for k=R;

enter image description here

  • $\begingroup$ Many thanks for pointing out this error, although I think you made one yourself as I think that the coefficient of the sine should now be $-1/2$. I wrote the limits correctly in my notes but typed them up wrong for some reason. I did actually use a similar convolution on a more difficult problem so I maybe should have spotted that there is a convolution here as well. Did you have any advice on how I might use this result to deduce the second integral? $\endgroup$
    – Tom
    Mar 11, 2018 at 17:22
  • $\begingroup$ "Did you have any advice on how I might use this result to deduce the second integral?" Which one are you speaking about ? It depends very much on $f$: for some simple $f$ one can obtain explicit expressions ; rather often, convolution products haven't an explicit expression, but nevertheless one can operate operations on them, in particular differentiation, due to the important result $(f \star g)'=f' \star g$ ($g$ doesn't need to be differentiable...). $\endgroup$
    – Jean Marie
    Mar 13, 2018 at 7:56
  • $\begingroup$ The integral I am looking for is $\int_{0}^{t}\tau\sin(2(t-\tau))$. Obviously I see that f in this case is just tau and so can't be much simpler, but I'm not entirely sure how I am supposed to use the result for the Laplace transform, is it related to differentiation under an integral? $\endgroup$
    – Tom
    Mar 13, 2018 at 9:28
  • $\begingroup$ Do I need to substitute this as a solution into the original differential equation and then use the fact that you have stated about differentiating convolution products to simplify? $\endgroup$
    – Tom
    Mar 13, 2018 at 9:38
  • $\begingroup$ Your first question : The result of the computation of $\int_{0}^{t}\tau\sin(2(t-\tau))d\tau$ is $\tfrac12(t-\tfrac12 \sin(2t))$. No need to differentiate under the integral sign (which is equivalent to the rule ($(f \star g)'=f' \star g$)). Your second question : if $f$ is more complicated than that, resisting to yield an exact expression, is there a general way to tackle convolutions ? I am tempted to answer : consider numerical methods... $\endgroup$
    – Jean Marie
    Mar 13, 2018 at 9:48

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