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Introduction to Differential Equations

Section 3.2 Transforms of derivatives and ODEs

Subsection 3.2.1 Transforms of derivatives

The Laplace Transform has a lot properties that mean it behaves nicely. In this video we’ll explore three crucial ones: linearity, existence, and inverses.
Correction: The Laplace transform of derivatives is missing some negative signs. Correct expressions for the Laplace of the first, second and third derivatives, respectively are: \[L[f’(t)]=sF(s)−f(0)\] \[L[f’’(t)]=s2F(s)−sf(0)−f’′(0),\] \[L[f’’’(t)]=s3F(s)−s2f(0)−sf’(0)−f’’(0)\]
Let us see how the Laplace transform is used for differential equations. First let us try to find the Laplace transform of a function that is a derivative. Suppose \(g(t)\) is a differentiable function of exponential order, that is, \(\lvert g(t) \rvert \leq M e^{ct}\) for some \(M\) and \(c\text{.}\) So \(\mathcal{L} \bigl\{ g(t) \bigr\}\) exists, and what is more, \(\lim_{t\to\infty} e^{-st}g(t) = 0\) when \(s > c\text{.}\) Then
\begin{equation*} \mathcal{L} \bigl\{ g'(t) \bigr\} = \int_0^\infty e^{-st} g'(t) \,dt = \Bigl[e^{-st} g(t) \Bigr]_{t=0}^\infty - \int_0^\infty (-s)\, e^{-st} g(t) \,dt = -g(0) + s \mathcal{L} \bigl\{ g(t) \bigr\} . \end{equation*}
We repeat this procedure for higher derivatives. The results are listed in Table 3.2.1. The procedure also works for piecewise smooth functions, that is functions that are piecewise continuous with a piecewise continuous derivative.
Table 3.2.1. Laplace transforms of derivatives (\(G(s) = \mathcal{L} \bigl\{ g(t) \bigr\}\) as usual).
\(f(t)\)   \(\mathcal{L} \bigl\{ f(t) \bigr\} = F(s)\)
\(g'(t)\)   \(sG(s)-g(0)\)
\(g''(t)\)   \(s^2G(s)-sg(0)-g'(0)\)
\(g'''(t)\)   \(s^3G(s)-s^2g(0)-sg'(0)-g''(0)\)

Subsection 3.2.2 Solving ODEs with the Laplace transform

Notice that the Laplace transform turns differentiation into multiplication by \(s\text{.}\) Let us see how to apply this fact to differential equations.
We finally know enough to go from start to finish. That is, we begin with an ODE. We transform it via the Laplace Transform. We manipulate it over in the world of Laplace Transforms. Then we translate it back to get a solution. Cool!

Example 3.2.2.

Take the equation
\begin{equation*} x''(t) + x(t) = \cos (2t), \quad x(0) = 0, \quad x'(0) = 1 . \end{equation*}
We will take the Laplace transform of both sides. By \(X(s)\) we will, as usual, denote the Laplace transform of \(x(t)\text{.}\)
\begin{equation*} \begin{aligned} \mathcal{L} \bigl\{ x''(t) + x(t) \bigr\} & = \mathcal{L} \bigl\{ \cos (2t) \bigr\} , \\ s^2 X(s) -sx(0)-x'(0) + X(s) & = \frac{s}{s^2 + 4} . \end{aligned} \end{equation*}
We plug in the initial conditions now—this makes the computations more streamlined—to obtain
\begin{equation*} s^2 X(s) -1 + X(s) = \frac{s}{s^2 + 4} . \end{equation*}
We solve for \(X(s)\text{,}\)
\begin{equation*} X(s) = \frac{s}{(s^2+1)(s^2 + 4)} + \frac{1}{s^2+1} . \end{equation*}
We use partial fractions (exercise) to write
\begin{equation*} X(s) =\frac{1}{3} \, \frac{s}{s^2+1} - \frac{1}{3}\, \frac{s}{s^2+4} + \frac{1}{s^2+1} . \end{equation*}
Now take the inverse Laplace transform to obtain
\begin{equation*} x(t) =\frac{1}{3} \cos (t) - \frac{1}{3} \cos (2t) + \sin (t) . \end{equation*}
The procedure for linear constant coefficient equations is as follows. We take an ordinary differential equation in the time variable \(t\text{.}\) We apply the Laplace transform to transform the equation into an algebraic (non differential) equation in the frequency domain. All the \(x(t)\text{,}\) \(x'(t)\text{,}\) \(x''(t)\text{,}\) and so on, will be converted to \(X(s)\text{,}\) \(sX(s) - x(0)\text{,}\) \(s^2X(s) - sx(0) - x'(0)\text{,}\) and so on. We solve the equation for \(X(s)\text{.}\) Then taking the inverse transform, if possible, we find \(x(t)\text{.}\)
It should be noted that since not every function has a Laplace transform, not every equation can be solved in this manner. Also if the equation is not a linear constant coefficient ODE, then by applying the Laplace transform we may not obtain an algebraic equation.
In this video we put everything together. We start with an IVP we take the Laplace transform of. The solution to the algebraic equation has a partial fraction decomposition with multiple types of factors and we finally take the inverse Laplace Transform to get a solution to the original IVP.

Subsection 3.2.3 Using the Heaviside function

A major strength of Laplace Transforms is in how it deals with discontinuities whether piecewise defined functions but even more so periodic functions that repeat the same discontinuous pattern over and over. The Laplace Transform converts these into continuous functions that can often be easier to work with
Before we move on to more general equations than those we could solve before, we want to consider the Heaviside function. See Figure 3.1 for the graph.
\begin{equation*} u(t) = \begin{cases} 0 & \text{if } \; t < 0 , \\ 1 & \text{if } \; t \geq 0 . \end{cases} \end{equation*}
Figure 3.1. Plot of the Heaviside (unit step) function \(u(t)\text{.}\)
This function is useful for putting together functions, or cutting functions off. Most commonly it is used as \(u(t-a)\) for some constant \(a\text{.}\) This just shifts the graph to the right by \(a\text{.}\) That is, it is a function that is 0 when \(t < a\) and 1 when \(t \geq a\text{.}\) Suppose for example that \(f(t)\) is a “signal” and you started receiving the signal \(\sin t\) at time \(t=\pi\text{.}\) The function \(f(t)\) should then be defined as
\begin{equation*} f(t) = \begin{cases} 0 & \text{if } \; t < \pi , \\ \sin t & \text{if } \; t \geq \pi . \end{cases} \end{equation*}
Using the Heaviside function, \(f(t)\) can be written as
\begin{equation*} f(t) = u(t - \pi) \, \sin t . \end{equation*}
Similarly the step function that is 1 on the interval \([1,2)\) and zero everywhere else can be written as
\begin{equation*} u(t - 1) - u(t-2) . \end{equation*}
The Heaviside function is useful to define functions defined piecewise. If you want to define \(f(t)\) such that \(f(t) = t\) when \(t\) is in \([0,1]\text{,}\) \(f(t) = -t+2\) when \(t\) is in \([1,2]\text{,}\) and \(f(t) = 0\) otherwise, then you can use the expression
\begin{equation*} f(t) = t \, \bigl( u(t) - u(t-1) \bigr) + (-t+2) \, \bigl( u(t - 1) - u(t-2) \bigr) . \end{equation*}
Hence it is useful to know how the Heaviside function interacts with the Laplace transform. We have already seen that
\begin{equation*} \mathcal{L} \bigl\{ u(t-a) \bigr\} = \frac{e^{-as}}{s} . \end{equation*}
This can be generalized into a shifting property or second shifting property.
\begin{equation} \mybxbg{~~ \mathcal{L} \bigl\{ f(t-a) \, u(t-a) \bigr\} = e^{-as} \mathcal{L} \bigl\{ f(t) \bigr\} . ~~}\tag{3.1} \end{equation}

Example 3.2.3.

The forcing function in our setup need not be periodic. Consider the mass-spring system
\begin{equation*} x''(t) + x(t) = f(t) , \quad x(0) = 0, \quad x'(0) = 0, \end{equation*}
where \(f(t) = 1\) if \(1 \leq t < 5\) and zero otherwise. Imagine a rocket attached to the mass is fired for 4 seconds starting at \(t=1\text{.}\) Or perhaps imagine an RLC circuit, where the voltage is raised at a constant rate for 4 seconds starting at \(t=1\text{,}\) and then held steady again starting at \(t=5\text{.}\)
We can write \(f(t) = u(t-1) - u(t-5)\text{.}\) We transform the equation and we plug in the initial conditions as before to obtain
\begin{equation*} s^2 X(s) + X(s) = \frac{e^{-s}}{s} - \frac{e^{-5s}}{s} . \end{equation*}
We solve for \(X(s)\) to obtain
\begin{equation*} X(s) = \frac{e^{-s}}{s(s^2+1)} - \frac{e^{-5s}}{s(s^2+1)} . \end{equation*}
We leave it as an exercise to the reader to show that
\begin{equation*} {\mathcal{L}}^{-1} \left\{ \frac{1}{s(s^2+1)} \right\} = 1 - \cos t . \end{equation*}
In other words \(\mathcal{L} \{ 1 - \cos t \} = \frac{1}{s(s^2+1)}\text{.}\) So using (3.1) we find
\begin{equation*} {\mathcal{L}}^{-1} \left\{ \frac{e^{-s}}{s(s^2+1)} \right\} = {\mathcal{L}}^{-1} \left\{ e^{-s} \mathcal{L} \{ 1 - \cos t \} \right\} = \bigl( 1 - \cos (t-1) \bigr) \, u(t-1) . \end{equation*}
\begin{equation*} {\mathcal{L}}^{-1} \left\{ \frac{e^{-5s}}{s(s^2+1)} \right\} = {\mathcal{L}}^{-1} \left\{ e^{-5s} \mathcal{L} \{ 1 - \cos t \} \right\} = \bigl( 1 - \cos (t-5) \bigr) \, u(t-5) . \end{equation*}
Hence, the solution is
\begin{equation*} x(t) = \bigl( 1 - \cos (t-1) \bigr) \, u(t-1) - \bigl( 1 - \cos (t-5) \bigr) \, u(t-5) . \end{equation*}
The plot of this solution is given in Figure 3.2.
Figure 3.2. Plot of \(x(t)\text{.}\)
This video is a start to finish walkthrough of the process of going from an IVP, converting via Laplace, dealing with the piecewise part, and then converting back to a solution to the original IVP. This video thus includes many pieces we’ve been developing so is probably your best check for comprehension yet. Feel free to try and work it out on your own first and skip to the end to check the solution.
TYPO: At about 4:00, in the red, \(\mathcal{L}\{u(t-a)f(t-a)\}=e^{-as}\mathcal{L}\{f(t)\}\text{.}\) I wrote the exponential with a t not an s. This typo persists in a few places in the video

Subsection 3.2.4 Transfer functions

The Laplace transform leads to the following useful concept for studying the steady state behavior of a linear system. Consider an equation of the form
\begin{equation*} L x = f(t) , \end{equation*}
where \(L\) is a linear constant coefficient differential operator. Then \(f(t)\) is usually thought of as input of the system and \(x(t)\) is thought of as the output of the system. For example, for a mass-spring system the input is the forcing function and the output is the behavior of the mass. We would like to have a convenient way to study the behavior of the system for different inputs.
Let us suppose that all the initial conditions are zero and take the Laplace transform of the equation, we obtain the equation
\begin{equation*} A(s) X(s) = F(s) . \end{equation*}
Solving for the ratio \(\nicefrac{X(s)}{F(s)}\) we obtain the so-called transfer function \(H(s) = \nicefrac{1}{A(s)}\text{,}\) that is,
\begin{equation*} H(s) = \frac{X(s)}{F(s)} . \end{equation*}
In other words, \(X(s) = H(s) F(s)\text{.}\) We obtain an algebraic dependence of the output of the system based on the input. We can now easily study the steady state behavior of the system given different inputs by simply multiplying by the transfer function.

Example 3.2.4.

Given \(x'' + \omega_0^2 x = f(t)\text{,}\) let us find the transfer function (assuming the initial conditions are zero).
First, we take the Laplace transform of the equation.
\begin{equation*} s^2 X(s) + \omega_0^2 X(s) = F(s) . \end{equation*}
Now we solve for the transfer function \(\nicefrac{X(s)}{F(s)}\text{.}\)
\begin{equation*} H(s) = \frac{X(s)}{F(s)} = \frac{1}{s^2 + \omega_0^2} . \end{equation*}
Let us see how to use the transfer function. Suppose we have the constant input \(f(t) = 1\text{.}\) Hence \(F(s) = \nicefrac{1}{s}\text{,}\) and
\begin{equation*} X(s) = H(s) F(s) = \frac{1}{s^2+\omega_0^2} \frac{1}{s} . \end{equation*}
Taking the inverse Laplace transform of \(X(s)\) we obtain
\begin{equation*} x(t) = \frac{1-\cos(\omega_0 t)}{\omega_0^2} . \end{equation*}

Subsection 3.2.5 Transforms of integrals

A feature of Laplace transforms is that it is also able to easily deal with integral equations. That is, equations in which integrals rather than derivatives of functions appear. The basic property, which can be proved by applying the definition and doing integration by parts, is
\begin{equation*} \mybxbg{~~ \mathcal{L} \left\{ \int_0^t f(\tau) ~ d\tau \right\} = \frac{1}{s} \, F(s) . ~~} \end{equation*}
It is sometimes useful (e.g. for computing the inverse transform) to write this as
\begin{equation*} \int_0^t f(\tau) ~ d\tau = {\mathcal{L}}^{-1} \left\{ \frac{1}{s} \, F(s) \right\} . \end{equation*}

Example 3.2.5.

To compute \({\mathcal{L}}^{-1} \left\{\frac{1}{s(s^2+1)}\right\}\) we could proceed by applying this integration rule.
\begin{equation*} {\mathcal{L}}^{-1} \left\{ \frac{1}{s} \, \frac{1}{s^2+1} \right\} = \int_0^t {\mathcal{L}}^{-1} \left\{ \frac{1}{s^2+1} \right\} ~ d\tau = \int_0^t \sin \tau ~ d\tau = 1 - \cos t . \end{equation*}

Example 3.2.6.

An equation containing an integral of the unknown function is called an integral equation. Consider
\begin{equation*} t^2 = \int_0^t e^{\tau} x(\tau) ~d\tau , \end{equation*}
where we wish to solve for \(x(t)\text{.}\) We apply the Laplace transform and the shifting property to get
\begin{equation*} \frac{2}{s^3} = \frac{1}{s} \, \mathcal{L} \bigl\{ e^{t} x(t) \bigr\} = \frac{1}{s} \, X(s-1) , \end{equation*}
where \(X(s) = \mathcal{L} \bigl\{ x(t) \bigr\}\text{.}\) Thus
\begin{equation*} X(s-1) = \frac{2}{s^2} \qquad \text{or} \qquad X(s) = \frac{2}{{(s+1)}^2}. \end{equation*}
We use the shifting property again
\begin{equation*} x(t) = 2 e^{-t} t . \end{equation*}

Subsection 3.2.6 Exercises

Exercise 3.2.2.

Compute the inverse Laplace transform of the function \(F(s)=\frac{1}{s^{2}\left(s-10\right)}\)
First we write \(F(s)=\frac{1}{s}\frac{1}{s\left(s-10\right)}\) and figure out the inverse Laplace transform of \(G(s)=\frac{1}{s\left(s-10\right)}\text{.}\) Using the property for Laplace transform of integrals, we write:
\begin{equation} \mathcal{L}^{-1}\left(\frac{1}{s\left(s-10\right)}\right)=\int_{0}^{t}\mathcal{L}^{-1}\left(\frac{1}{s-10}\right)d\tau=\int_{0}^{t}e^{10\tau}d\tau=\frac{1}{10}\left(e^{10t}-1\right)\nonumber\tag{3.2} \end{equation}
Now we write
\begin{equation} \mathcal{L}^{-1}\left(\frac{1}{s^{2}\left(s-10\right)}\right)=\int_{0}^{t}\mathcal{L}^{-1}\left(\frac{1}{s\left(s-10\right)}\right)d\tau=\int_{0}^{t}[e^{10\tau}-1] d\tau=\frac{1}{100} \left(-10 t+e^{10 t}-1\right)\nonumber\tag{3.3} \end{equation}

Exercise 3.2.3.

Compute the inverse Laplace transform of the function \(F(s)=\frac{1}{s^{2}\left(s^{2}+1\right)}\)
Using the example from section above, we know that \(\mathcal{L}^{-1}\left(\frac{1}{s\left(s^{2}+1\right)}\right)=1-\cos t\text{.}\) Using this and the property for Laplace transform of integrals, we write:
\begin{equation} \mathcal{L}^{-1}\left(\frac{1}{s^{2}\left(s^{2}+1\right)}\right)=\mathcal{L}^{-1}\left(\frac{1}{s}\frac{1}{s\left(s^{2}+1\right)}\right)=\int_{0}^{t}\mathcal{L}^{-1}\left(\frac{1}{s\left(s^{2}+1\right)}\right)d\tau=\int_{0}^{t}(1-\cos\tau)d\tau=t-\sin t\nonumber\tag{3.4} \end{equation}

Exercise 3.2.4.

Using the Laplace transform solve
\begin{equation*} m x'' + c x' + k x = 0 , \quad x(0) = a, \quad x'(0) = b , \end{equation*}
where \(m > 0\text{,}\) \(c > 0\text{,}\) \(k > 0\text{,}\) and \(c^2 - 4km > 0\) (system is overdamped).
Taking the Laplace transform
\begin{equation*} \begin{aligned} m\left(s^2X-sa-b\right)+c\left(sX-a\right)+kX=0 \end{aligned} \end{equation*}
Solving for \(X\)
\begin{equation*} \begin{aligned} X=\frac{mas+mb+ca}{ms^2+cs+k}=\frac{mas}{ms^2+cs+k}+\frac{mb+ca}{ms^2+cs+k} \end{aligned} \end{equation*}
Completing the square and noting that \(4km-c^2<0\)
\begin{equation*} \begin{aligned} X=a\frac{s}{(s+\frac{c}{2m})^2-\frac{1}{4m^2}(c^2-4km)}+\frac{b+\frac{ca}{m}}{\frac{1}{2m}\sqrt{c^2-4km}}\frac{\overbrace{\frac{1}{2m}\sqrt{c^2-4km}}^{\omega}}{(s+\frac{c}{2m})^2-\frac{1}{4m^2}(c^2-4km)} \end{aligned} \end{equation*}
Noting the shift \(+\frac{c}{2m}\text{,}\) we rewrite
\begin{equation*} \begin{aligned} X=a\frac{s+\frac{c}{2m}}{(s+\frac{c}{2m})^2-\frac{1}{4m^2}(c^2-4km)}+\frac{b+\frac{ca}{2m}}{\frac{1}{2m}\sqrt{c^2-4km}}\frac{\frac{1}{2m}\sqrt{c^2-4km}}{(s+\frac{c}{2m})^2-\frac{1}{4m^2}(c^2-4km)} \end{aligned} \end{equation*}
Now taking the inverse Laplace transform we get
\begin{equation*} \begin{aligned} x(t)=ae^{-\frac{c}{2m}t}\sinh\left(\frac{\sqrt{c^2-4km}}{2m}t\right)+\frac{2mb+ca}{\sqrt{c^2-4km}}e^{-\frac{c}{2m}t}\cosh\left(\frac{\sqrt{c^2-4km}}{2m}t\right) \end{aligned} \end{equation*}

Exercise 3.2.5.

Using the Laplace transform solve
\begin{equation*} m x'' + c x' + k x = 0 , \quad x(0) = a, \quad x'(0) = b , \end{equation*}
where \(m > 0\text{,}\) \(c > 0\text{,}\) \(k > 0\text{,}\) and \(c^2 - 4km < 0\) (system is underdamped).

Exercise 3.2.6.

Using the Laplace transform solve
\begin{equation*} m x'' + c x' + k x = 0 , \quad x(0) = a, \quad x'(0) = b , \end{equation*}
where \(m > 0\text{,}\) \(c > 0\text{,}\) \(k > 0\text{,}\) and \(c^2 = 4km\) (system is critically damped).

Exercise 3.2.7.

Using the Heaviside function write down the piecewise function that is 0 for \(t < 0\text{,}\) \(t^2\) for \(t\) in \([0,1]\) and \(t\) for \(t > 1\text{.}\)

Exercise 3.2.8.

Solve \(x'' + x = u(t-1)\) for initial conditions \(x(0) = 0\) and \(x'(0) = 0\text{.}\)
Taking the Laplace transform
\begin{equation*} \begin{aligned} s^2X+X= &\frac{e^{-s}}{s} \\ X=&\frac{e^{-s}}{(s^2+1)s} = e^{-s}\left(\frac{1}{s}-\frac{s}{s^2+1}\right) \end{aligned} \end{equation*}
Taking the inverse Laplace transform
\begin{equation*} \begin{aligned} x(t)=\left(1-\cos(t-1)\right)u(t-1) \end{aligned} \end{equation*}

Exercise 3.2.9.

Solve \(x''' + x = (t-1)^3 u(t-1)\) for initial conditions \(x(0) = 1\) and \(x'(0) = 0\text{,}\) \(x''(0) = 0\text{.}\)

Exercise 3.2.10.

Solve \(x''-x = (t^2-1) u(t-1)\) for initial conditions \(x(0)=1\text{,}\) \(x'(0) = 2\) using the Laplace transform.
\(x(t) = (2e^{t-1}-t^2-1) u(t-1) -\frac{1}{2}e^{-t}+\frac{3}{2}e^t\)

Exercise 3.2.11.

Show the second shifting property: \(\mathcal{L} \bigl\{ f(t-a) \, u(t-a) \bigr\} = e^{-as} \mathcal{L} \bigl\{ f(t) \bigr\}\text{.}\)
Using the definition
\begin{equation*} \begin{aligned} \mathcal{L} \bigl\{ f(t-a)u(t-a) \bigr\} =&\int_0^\infty e^{-st}f(t-a)u(t-a)dt \\ =& \int_a^\infty e^{-st}f(t-a)dt \end{aligned} \end{equation*}
Defining \(\tau\equiv t-a\) we get
\begin{equation*} \begin{aligned} \int_0^\infty e^{-as}e^{-s\tau}f(\tau)d\tau=e^{-as}\underbrace{\int_0^\infty e^{-s\tau}f(\tau)}_{F(s)}d\tau \end{aligned} \end{equation*}
\begin{equation*} \begin{aligned} \mathcal{L} \bigl\{ f(t-a)u(t-a) \bigr\} = e^{-as}\mathcal{L}\bigl\{ f(t) \bigr\} \end{aligned} \end{equation*}

Exercise 3.2.12.

Let us think of the mass-spring system with a rocket from Example 3.2.3. We noticed that the solution kept oscillating after the rocket stopped running. The amplitude of the oscillation depends on the time that the rocket was fired (for 4 seconds in the example).
  1. Find a formula for the amplitude of the resulting oscillation in terms of the amount of time the rocket is fired.
  2. Is there a nonzero time (if so what is it?) for which the rocket fires and the resulting oscillation has amplitude 0 (the mass is not moving)?
a) From the example
\begin{equation*} \begin{aligned} x(t)=(1-\cos(t-1))u(t-1)-(1-\cos(t-5))u(t-5) \end{aligned} \end{equation*}
At \(t>5\)
\begin{equation*} \begin{aligned} x(t)=& \cos(t-5)-\cos(t-1) \\ =& \underbrace{2\sin\left(\frac{5-1}{2} \right)}_{A}\sin\left( t-\frac{5+1}{2}\right) \end{aligned} \end{equation*}
Where \(|A|\approx 1.81\) is the amplitude. Defining \(t_f\) as the time the rocket stops firing, and \(t_0=t_f-1\) as the duration the rocket fires, we can generalize this result
\begin{equation*} \begin{aligned} x(t)=2\sin\left( \frac{t_0}{2}\right)\sin\left( t-\frac{t_0+2}{2} \right) \end{aligned} \end{equation*}
So the amplitude is given by
\begin{equation*} \begin{aligned} A(t_0)=2\lvert \sin(t_0/2)\rvert \end{aligned} \end{equation*}

b) From \(A(t_0)=2\lvert \sin(t_0/2)\rvert\) this is zero when \(t_0=2n\pi\) for \(n\) a positive integer.

Exercise 3.2.13.

\begin{equation*} f(t) = \begin{cases} {(t-1)}^2 & \text{if } \; 1 \leq t < 2, \\ 3-t & \text{if } \; 2 \leq t < 3, \\ 0 & \text{otherwise} . \end{cases} \end{equation*}
  1. Sketch the graph of \(f(t)\text{.}\)
  2. Write down \(f(t)\) using the Heaviside function.
  3. Solve \(x''+x=f(t)\text{,}\) \(x(0)=0\text{,}\) \(x'(0) = 0\) using Laplace transform.
b) \(u(t-1)(t-1)^2+u(t-2)\left((3-t)-(t-1)^2\right)-u(t-3)(t-3)\)
c) \(x(t)=u(t-1)\bigl[(t-1)^2-2+2\cos(t-1)\bigr]+u(t-2)\bigl[-t^2+t+4+3\sin(t-2)-2\cos(t-2)\bigr]-u(t-3)\bigl[(t-3)-\sin(t-3)\bigr]\)

Exercise 3.2.14.

Using the Heaviside function \(u(t)\text{,}\) write down the function
\begin{equation*} f(t) = \begin{cases} 0 & \text{if } \; \phantom{1 \leq {}} t < 1 , \\ t-1 & \text{if } \; 1 \leq t < 2 , \\ 1 & \text{if } \; 2 \leq t . \end{cases} \end{equation*}
\(f(t) = (t-1)\bigl(u(t-1) - u(t-2)\bigr) + u(t-2)\)

Exercise 3.2.15.

Find the transfer function for \(x' + x = f(t)\) (assuming the initial conditions are zero).
\(H(s) = \frac{1}{s+1}\)

Exercise 3.2.16.

Find the transfer function for \(m x'' + c x' + kx = f(t)\) (assuming the initial conditions are zero).
Taking the Laplace transform
\begin{equation*} \begin{aligned} ms^2X+csX+kX=F(s) \end{aligned} \end{equation*}
The transfer function is
\begin{equation*} \begin{aligned} H(s)=\frac{X}{F}=\frac{1}{ms^2+cs+k} \end{aligned} \end{equation*}

Exercise 3.2.17.

Show the differentiation of the transform property. Suppose \(\mathcal{L} \bigl\{ f(t) \bigr\} = F(s)\text{,}\) then show
\begin{equation*} \mathcal{L} \bigl\{ -t f(t) \bigr\} = F'(s) . \end{equation*}
Hint: Differentiate under the integral sign.
\begin{equation*} \begin{aligned} F'(s)=&\frac{dF(s)}{ds}=\frac{d}{ds}\int_0^\infty f(t)e^{-st}dt=\int_0^\infty f(t)\frac{d}{ds}e^{-st}dt=\int_0^\infty \underbrace{\left[ -tf(t) \right]}_{\textrm{call this } g(t)}e^{-st}dt \\ =& \int_0^\infty g(t) e^{-st} dt=\mathcal{L}\{ g(t)\}=\mathcal{L} \{ -tf(t)\} \end{aligned} \end{equation*}

Exercise 3.2.18.

Find the Laplace transform of \(f(t)=t\sin(3t)\) (assuming the initial conditions are zero).
\begin{equation} \mathcal{L}[f(t)]=-\frac{d}{ds}\mathcal{L}[\sin(3t)]=-\frac{d}{ds}\left[\frac{3}{s^{2}+9}\right]=\frac{6s}{(s^{2}+9)^{2}}\nonumber\tag{3.5} \end{equation}

Exercise 3.2.19.

Find the Laplace transform of \(f(t)=te^{t}\sin(3t)\) (assuming the initial conditions are zero).
\begin{equation} \mathcal{L}[f(t)]=-\frac{d}{ds}\mathcal{L}[e^{t}\sin(3t)]=-\frac{d}{ds}\left[\frac{3}{(s-1)^{2}+9}\right]=\frac{6 (s-1)}{\left(s^2-2 s+10\right)^2}\nonumber\tag{3.6} \end{equation}

Exercise 3.2.20.

Solve the following initial value problem: \(tx''+x'=t^{3}\) with \(x(0)=0\text{.}\)
Taking the Laplace transform of the given differential equation, we get
\begin{equation} -\frac{d}{ds}\left[s^{2}X(s)-sx(0)-x'(0)\right]+sX(s)-x(0)=\frac{6}{s^{4}}\nonumber\tag{3.7} \end{equation}
This becomes \(X'(s)+\frac{1}{s}X(s)=-\frac{6}{s^{6}}\text{.}\) Solving this, we get \(X(s)=\frac{c}{s}+\frac{3}{2s^{5}}\text{.}\) Taking the inverse Laplace of this we get \(x(t)=C+\frac{t^{4}}{16}\text{.}\)
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