Solving O.D.E and Initial Values Problem using Laplace Transform

Question

I have this ODE:

$$ y'' + y = \begin{cases} \cos t, &\text{ if }0\le t \lt \pi\\ t-\pi,&\text{ if }\pi \le t \lt \infty \end{cases} $$

The initial values are: $$ y(0)=0 \\ y'(0)=0 $$

I want to solve it using Laplace and reverse Laplace transform.

I know that the left part of the equation will become:

$$ s^2L(y) - sy(0) -y'(0)+ L(y) $$ and because of the inital values are zeroed out, we are left with:

$$ s^2L(y) + L(y) $$

For the right part, i will have to break it into two integrals. The one will be from $$ [0,π] $$

and the other one will be from $$ [π, \infty] $$

Finally, i know from the Laplacian identities that the Laplace transform of $$ cos(at) = \frac{s}{s^2 + a^2} $$

So what is left is to compute the Laplace transform of $$ t-π $$

Unfortunately at this point i am stuck. Even if i found the answer somewhere, i am afraid that i might be stuck at the reversal of the procedure, in order to calculate the inverse Laplace transform of

$$ t-π $$

Answer 1

You are correct that you should break the right-hand-side into two pieces, but you will end up with (RHS=Right Hand Side) $$ \int_0^{+\infty}e^{-st}\cdot \text{RHS}\,dt= \int_0^\pi e^{-st}\cos t\,dt+\int_\pi^{+\infty}e^{-st}(t-\pi)\,dt. $$ Then just evaluate those integrals with your favorite method (mine would be integrating by parts).

Answer 2

The trick is to solve for $t$ in the region $t < \pi$ using Laplace transforms. Given the initial conditions, this gives you an unambiguous solution, from which you can read off $y(\pi)$ and $y'(\pi)$.

Then you can easily solve the initial value problem $$y'' + y = t-\pi$$ starting at the now-known $y(\pi)$ and $y'(\pi)$.

Answer 3

$\int_0^\pi e^{-st}\cos t~dt$ is not the Laplace transform of $\cos t$, which actually is $\int_0^\infty e^{-st}\cos t~dt=\frac s{s^2+1}$. You can easily solve this by repeated integration by parts. Or you could write $\cos t=\mathfrak R(e^{it})$ and hence,$$\int_0^\pi e^{-st}\mathfrak R(e^{it})~dt=\mathfrak R\left(\int_0^\pi e^{(i-s)t}~dt\right)\\=\mathfrak R\left(\frac1{i-s}[e^{-\pi s}e^{i\pi}-1]\right)=\frac{s(e^{-\pi s}+1)}{s^2+1}$$

For the second term $\int_\pi^\infty e^{-st}(t-\pi)~dt$, you may substitute $m=t-\pi$,$$e^{-\pi s}\int_0^\infty e^{-sm}m~dm=e^{-\pi s}\mathcal L[t]=\frac{e^{-\pi s}}{s^2}$$Alternatively, you may note that $\int_\pi^\infty e^{-st}(t-\pi)~dt=\mathcal L[f(t-\pi)H(t-\pi)]$, where $f(t)=t$ and $H(t)$ is the unit-step function. This is a well known formula in Laplace transform and evaluates to $e^{-\pi s}\mathcal L[f(t)]=e^{-\pi s}\mathcal L[t]=\frac{e^{-\pi s}}{s^2}$.

Answer 4

You may refer to the time-shifting theorem, which states

$$ \mathcal L\{ u(t-a)f(t-a) \} = e^{-as}F(s) $$

where $u(t)$ is the Heaviside step function. This means, the Laplace transform of $t-\pi$ for $t>\pi$ is the Laplace transform of $f(t)=t$ for $t>0$, multiplied with a factor of $e^{-\pi s}$. You should already know that

$$ F(s) = \mathcal L\{t\} = \frac{1}{s^2} $$

Hence, the result turns out to be

$$ \mathcal L\{ u(t-\pi)(t-\pi) \} = \frac{e^{-\pi s}}{s^2} $$

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This isn't over yet, as still have to transform the remaining piece. Note that it is not $\frac{s}{1+s^2}$, since that's the transform for $\cos t$ over all $t>0$.

You can rewrite the RHS function as

$$ g(t) = \cos t \big[1 - u(t-\pi)\big] + (t-\pi)u(t-\pi) $$

The second term has been transformed as above. For the first term, you can use the same shifting theorem as above, noting that

$$ \cos (t) \big[1 - u(t-\pi)\big] = \cos (t) - \cos (t) u(t-\pi) = \cos t + \cos(t-\pi)u(t-\pi) $$

So the final transform is

$$ G(s) = \frac{s}{1+s^2} + \frac{e^{-\pi s}s}{1+s^2} + \frac{e^{-\pi s}}{s^2} $$

To find the solution, you must use the shifting theorem in reverse (after applying partial fractions to the transformed solution)