Section 6.5 ODEs with discontinuous forcing functions
We will now do some examples involving intial value problems where the forcing function have pieces that switch on and off. This will require the use of step functions to turn parts of the forcing function on and off as the definition requires. You can think of this as a way to rewrite a multi-part definition into just one definition for a piecewise continuous function.Example 6.5.1.
Solve using Laplace Transforms:
Step 1: Take \(\mathcal{L}\) of both sides and solve for \(\mathcal{L}\)
so that
Step 2: Solve for \(\mathcal{L}\left\{ y\right\} \text{,}\)
Step 3: We do partial fractions on
Step 4: Take the inverse Laplace transform: Using \(\mathcal{L}\left[u_{a}(t)f(t-a)\right]=e^{-as}F(s)\text{,}\) and get
Example 6.5.2.
Solve using Laplace Transforms:
Step 1: Take \(\mathcal{L}\) of both sides
and get
Step 2: Solve for \(\mathcal{L}\left\{ y\right\} \) and get
Step 3: We do partial fractions
Step 3: Take the inverse Laplace transform: Using \(\mathcal{L}\left[u_{a}(t)f(t-a)\right]=e^{-as}F(s)\text{,}\) and
Example 6.5.3.
Solve using Laplace Transforms:
Step 1: Take \(\mathcal{L}\) of both sides and solve for \(\mathcal{L}\)
and recall \(\mathcal{L}\left[u_{a}(t)f(t-a)\right]=e^{-as}F(s)\text{,}\) hence \(a=4\text{,}\) \(f(t-5)=\sin\left(t-5\right)\) hence \(f(t)=\sin t\) and \(\mathcal{L}\left\{ \sin t\right\} =\frac{1}{s^{2}+1}\) hence
Step 2: We do partial fractions on
hence
hence
and get
hence
Step 3: Take the inverse Laplace transform: Using \(\mathcal{L}\left[u_{a}(t)f(t-a)\right]=e^{-as}F(s)\text{,}\) and \(\mathcal{L}\left\{ \sin(at)\right\} =\frac{a}{s^{2}+a^{2}}\) and \(\mathcal{L}\left\{ \cos(at)\right\} =\frac{a}{s^{2}+a^{2}}\) we have
Example 6.5.4.
Solve using Laplace Transforms:
Step 1: Take \(\mathcal{L}\) of both sides and solve for \(\mathcal{L}\)
hence
Step 2: We do partial fractions on
hence
after doing the work to get the partial fractions you get
putting it back we need to take the inverse of
Step 3: Take the inverse Laplace transform: Using \(\mathcal{L}\left[u_{a}(t)f(t-a)\right]=e^{-as}F(s)\text{,}\) and \(\mathcal{L}\left\{ \cos(at)\right\} =\frac{a}{s^{2}+a^{2}}\) and \(\mathcal{L}\left\{ e^{at}\right\} =\frac{1}{s-a}\) we have
Example 6.5.5.
Find the Laplace transform of
Step 1: First let us rewrite this in terms of unit step functions. When \(0\leq t\lt 1\text{:}\) the function is \(f(t)=t\) When \(1\leq t\lt \infty\text{:}\) then function is \(f(t)=t+?\cdot u_{1}(t)=3t\) hence \(?=2t\) so that
Step 2: Before we can take a Laplace transform, we notice that our fomula involves \(\mathcal{L}\left\{ u_{c}(t)g\left(t-c\right)\right\} =e^{-ct}\mathcal{L}\left\{ g(t)\right\} \text{.}\) Thus we will need to turn \(2tu_{1}(t)\) into this form:
hence
Fact 6.5.6.
We have
Example 6.5.7.
Take the Laplace transform of \(f(t)=u_{1}(t)te^{t}\text{.}\)
Notice that we cannot use the formula \(\mathcal{L}\left\{ u_{c}(t)g\left(t-c\right)\right\} =e^{-ct}\mathcal{L}\left\{ g(t)\right\} \) directly since \(te^{t}\) is not written as a function of \((t-1)\text{.}\) Hence we'll need to use \(\mathcal{L}\left\{ u_{c}(t)h(t)\right\} =e^{-cs}\mathcal{L}\left\{ h\left(t+c\right)\right\} \) with \(h(t)=te^{t}\) and \(c=1\text{.}\) Thus
and we get
where we used formula \(2\) and \(11\) in the table.