Exercises – Long Day#
Exercise 1: Homogeneous or Inhomogeneous?#
Are the following differential equations and systems of differential equations homogeneous or inhomogeneous?
\(f''(t)=f'(t)-2f(t)\).
\(f'(t)-t\cdot f(t)-\mathrm e^{-3t}=0\).
\(\left[\begin{array}{c} f'_1(t)\\ f'_2(t) \end{array}\right] =\left[\begin{array}{cc} 4 & 6 \\ -2 & 7 \end{array}\right] \cdot \left[\begin{array}{c} f_1(t)\\ f_2(t) \end{array}\right]+\left[\begin{array}{c} 1\\ 1\end{array}\right].\)
\(\left\{ \begin{array}{lcr} f'_1(t) & = & f_2(t)\\ f'_2(t) & = & f_1(t)-f_2(t). \end{array} \right.\)
Answer
The equation can be rewritten to \(f''(t)-f'(t)+2f(t)=0\). The differential equation is thus homogeneous (see Definition 12.3.1 in the textbook).
The equation can be rewritten to \(f'(t)-t\cdot f(t)=\mathrm e^{-3t}\). The differential equation is thus inhomogeneous.
The system of differential equations is inhomogeneous (see Definition 12.2.1 in the textbook).
The system of differential equations can be rewritten to \(\left[\begin{array}{c} f'_1(t)\\ f'_2(t) \end{array}\right] =\left[\begin{array}{cc} 0 & 1\\1 & -1 \end{array}\right] \cdot \left[\begin{array}{c} f_1(t)\\ f_2(t) \end{array}\right].\) Hence the differential equation system is homogeneous.
Exercise 2: A Homogeneous Second-Order Differential Equation#
We are given the homogeneous real differential equation
Compute its general solution.
Hint
First find all roots of the characteristic polynomial of the equation. See Section 12.3 in the textbook for more information and examples.
Answer
Exercise 3: More Homogeneous Second-Order Differential Equations#
We are given the following homogeneous real differential equations. Compute their general solutions.
- \[f''(t)-6f'(t)+9f(t)=0.\]
- \[f''(t)+2f'(t)+5f(t)=0.\]
Hint
Depending on whether the characteristic polynomial has two real roots, two non-real roots, or a double-root, one must use Case 1, Case 2, or Case 3 on pages 266 and 267 in Section 12.3 in the textbook.
Answer
\(f(t)=c_1 \cdot \mathrm e^{3t}+c_2 \cdot t\mathrm e^{3t}, \quad c_1,c_2 \in \mathbb{R}.\)
\(f(t)=c_1 \cdot \mathrm e^{−t}\mathrm{cos}(2t)+c_2\cdot \mathrm e^{−t} \mathrm{sin}(2t), \quad c_1,c_2 \in \mathbb{R}.\)
Exercise 4: An Inhomogeneous Second-Order Differential Equation#
We are given the inhomogeneous differential equation
Question a#
Determine a real number \(a\) such that the function \(f(t)=a\cdot \mathrm e^{2t}\) is a particular solution to the given differential equation.
Hint
Insert the function \(f(t)=a\cdot \mathrm e^{2t}\) in the differential equation. Which equation must \(a\) fulfill so that the function is a solution to the differential equation?
Answer
The function \(f(t)=\mathrm e^{2t}\) is a particular solution.
Question b#
Now describe the general solution to the given inhomogeneous differential equation. Note that the corresponding homogeneous differential equation already has been treated in Exercise 3.
Hint
The general solution to the corresponding homogeneous differential equation \(f''(t)-6f'(t)+9f(t)=0\) was in Exercise 3 found to be \(f(t)=c_1 \cdot \mathrm e^{3t}+c_2 \cdot t\mathrm e^{3t}, \quad c_1,c_2 \in \mathbb{R}.\)
Answer
The wanted general solution is the sum of a particular solution and the general solution to the corresponding homogeneous differential equation. Hence the answer is:
Exercise 5: Rewriting of a Higher-Order Differential Equation to a System#
A linear second-order differential equation with constant coefficients can be rewritten to a system of first-order differential equations (see Section 12.3 in the textbook). In this exercise we consider an example.
Question a#
We are given the differential equation \(f''(t)+2f'(t)+f(t)=\mathrm{cos}(t)\). Rewrite this differential equation to a system of two first-order differential equations in the functions \(f_1(t)\) and \(f_2(t)\), where \(f_1(t)=f(t)\) and \(f_2(t)=f'(t)\).
Answer
If we choose \(f_1(t)=f(t)\) and \(f_2(t)=f'(t)\), then it applies that \(f'_1(t)=f'(t)=f_2(t)\) and \(f'_2(t)=f''(t)=-2f'(t)-f(t)+\mathrm{cos}(t)=-2f_2(t)-f_1(t)+\mathrm{cos}(t)\). Hence we get the differential equation system
Question b#
Check that Equation (12.13) would have given the same system.
Answer
If Equation (12.13) is used with \(a_0=1\), \(a_1=2\), and \(q(t)=\mathrm{cos}(t)\), then we get the same differential equation system as in the answer to Question a.
Exercise 6: One more Inhomogeneous Second-Order Differential Equation#
We are given the inhomogeneous differential equation
Question a#
Inspired by Exercise 4 one could hope that a real number \(a\) exists such that \(f(t)=a\cdot \mathrm e^t\) is a particular solution to the given differential equation. Show that in fact no function of the form \(f(t)=a\cdot \mathrm e^t\) is a solution. What is the problem?
Answer
Insertion of a function with the form \(f(t)=a \cdot \mathrm e^{t}\) into the differential equation results in \(0=\mathrm e^t\), which never is fulfilled. The problem is that functions of the form \(f(t)=a \cdot \mathrm e^{t}\) already is a solution to the corresponding homogeneous differential equation \(f''(t)+3f'(t)-4f(t)=0.\)
Question b#
Now try finding a particular solution of the form \(f(t)=a \cdot t\cdot \mathrm e^{t}\). Check first that \((t\cdot \mathrm e^{t})'=\mathrm e^t+t\cdot \mathrm e^{t}\) and \((t\cdot \mathrm e^{t})''=2\mathrm e^t+t\cdot \mathrm e^{t}\).
Answer
By inserting the function \(f(t)=a\cdot t\cdot \mathrm e^{t}\) into the differential equation you will, after a few computational steps, arrive at \(2a\mathrm e^t+3a\mathrm e^t=\mathrm e^t\). This equation is fulfilled precisely when \(a=1/5\). Hence the function \(f(t)=(1/5)t\mathrm e^t\) is a particular solution to the given differential equation.
Question c#
Now describe the general solution to the given inhomogeneous differential equation. Note that the corresponding homogeneous differential equation was investigated already in Exercise 2.
Hint
The general solution to the corresponding homogeneous differential equation \(f''(t)+3f'(t)-4f(t)=0\) was in Exercise 2 found to be \(f(t)=c_1 \cdot \mathrm e^{t}+c_2 \cdot \mathrm e^{-4t}, \quad c_1,c_2 \in \mathbb{R}.\)
Answer
The wanted general solution is the sum of a particular solution and the general solution to the corresponding homogeneous differential equation. Hence the answer is:
Exercise 7: From Solution to Differential Equation#
We are informed that the general real-valued solution to the homogeneous linear second-order differential equation with constant real coefficients is
Write up the differential equation.
Hint
Since \(0\) is a solution (let \(c_1=0\) and \(c_2=0\) in the general solution), then the wanted differential equation is homogeneous. The differential equation can thus be written in the form \(f''(t)+a_1 f'(t)+a_0 f(t)=0\) for some real numbers \(a_0,a_1\).
Hint
Which roots must the characteristic polynomial \(Z^2+a_1Z+a_0\) have?
Answer
From the given general solution one sees that the characteristic polynomial \(Z^2+a_1Z+a_0\) has roots \(-1+2i\) and \(-1-2i\). Thus it applies that \(Z^2+a_1Z+a_0=(Z-(-1+2i))\cdot (Z-(-1-2i))\). After expanding the brackets one gets \(Z^2+a_1Z+a_0=Z^2+2Z+5.\) Thus \(a_1=2\) and \(a_0=5\). The wanted differential equation is: \(f''(t)+2f'(t)+5f(t)=0.\)
Exercise 8: From Solution to Differential Equation, Part 2#
We are informed that the general, real-valued solution to an inhomogeneous, linear second-order differential equation is
Write out the differential equation.
Hint
What is the general solution to the corresponding homogeneous differential equation? Can the corresponding homogeneous differential equation be determine using the answer to Exercise 7?
Hint
The corresponding homogeneous differential equation is according to Exercise 7 given as \(f''(t)+2f'(t)+5f(t)=0.\) The inhomogeneous differential equation that we want to find now can thus be written in the form \(f''(t)+2f'(t)+5f(t)=q(t).\)
Hint
The function \(f(t)=7+3t+5\mathrm e^t\) is a particular solution to the wanted differential equation. What do you get if you insert this function in the latter differential equation in the previous hint?
Answer
The wanted differential equation is \(f''(t)+2f'(t)+5f(t)=41+15t+40\mathrm e^t.\)
Exercise 9: Initial Conditions#
We are given the homogeneous real differential equation
Note that the differential equation is the same as in Exercise 1. The goal of this exercise is to find the solution to the differential equation that fulfills the initial conditions \(f(0)=1\) and \(f'(0)=2\).
Question a#
In Exercise 1 the result was that the given differential equation has the general solution
Which equation must \(c_1\) and \(c_2\) fulfill in order for \(f(0)=1\) to apply?
Answer
From Exercise 1 we know that
Inserting \(t=0\) and using the initial condition \(f(0)=1\) results in \(1=c_1+c_2.\)
Question b#
Which equation must \(c_1\) and \(c_2\) fulfill so that \(f'(0)=2\) applies?
Hint
As in Question a we know from Exercise 1 that
Now compute the derivative of both sides of the equal sign and then insert \(t=0\).
Hint
From the general solution
we get from the differentiation on both sides of the equal side that
Now use the last part of the previous hint.
Answer
\(2=c_1-4c_2\)
Question c#
Now find the solution \(f(t)\) to the differential equation that fulfills \(f(0)=1\) and \(f'(0)=2\).
Hint
Compute \(c_1\) and \(c_2\) by solving the two equations that were found in Questions a and b.
Answer
Solving the two equations \(1=c_1+c_2\) and \(2=c_1-4c_2\) results in \(c_1=6/5\) and \(c_2=-1/5\). Hence the wanted solution is: