Exercises – Long Day

Exercises – Long Day#

Exercise 1: A Real, Non-Linear Differential Equation#

We are given the real differential equation \(f'(t)^2-4\mathrm e^{2t} \cdot f(t)=0\).

Question a#

Is the function \(f(t)=t\) a solution to the differential equation?

Hint

Is the equation \(f'(t)^2-4\mathrm e^{2t} \cdot f(t)= 0\) fulfilled if you insert the function \(f(t)=t\)?

Answer

No.

Question b#

Is the function \(f(t)=\mathrm e^{2t}\) a solution to the differential equation?

Hint

The derivative of \(f(t)=\mathrm e^{2t}\) is \(f'(t)=2\mathrm e^{2t}\). Is the equation \(f'(t)^2-4\mathrm e^{2t} \cdot f(t)= 0\) fulfilled if you insert the function \(f(t)=\mathrm e^{2t}\)?

Answer

Yes.

Exercise 2: A Real, Linear Differential Equation#

In this exercise we investigate the real differential equation

\[f'(t)-3f(t)=t.\]

Question a#

We are being informed that real numbers \(a\) and \(b\) exist such that the function \(at+b\) is a solution to the differential equation. Compute \(a\) and \(b\).

Hint

Insert the function \(f(t)=at+b\) into the differential equation and investigate what \(a\) and \(b\) must fulfill for the function to be a solution.

Answer

\(a=-1/3\) and \(b=-1/9\).

Question b#

Determine the general solution to the real differential equation \(f'(t)-3f(t)=0\).

Answer

\(f(t)=c \cdot \mathrm e^{3t}\), where \(c \in \mathbb{R}.\)

Question c#

Use your answers to the previous questions to find the general solution to the real differential equation \(f'(t)-3f(t)=t\).

Answer

The wanted general solution is \(f(t)=c \cdot \mathrm e^{3t}+(-1/3)t+(-1/9)\), where \(c \in \mathbb{R}.\) It is a bit simpler to write \(f(t)=c \cdot \mathrm e^{3t}-t/3-1/9\), where \(c \in \mathbb{R}.\)

Exercise 3: Initial Conditions#

We are being informed that the general solution to the real differential equation \(f'(t)=\mathrm e^t \cdot f(t)\) is given by \(f(t)=c\cdot \mathrm e^{\mathrm e^t},\) where \(c \in \mathbb{R}.\)

Question a#

Check by insertion into the given differential equation that \(f(t)=3\cdot \mathrm e^{\mathrm e^t}\) indeed is a solution.

Question b#

Find the solution to the differential equation that fulfills the initial-value condition \(f(0)=1\).

Hint

Since the general solution is given by \(f(t)=c\cdot \mathrm e^{\mathrm e^t},\) one must try to determine a value of \(c\) such that the initial-value condition is fulfilled.

Answer

The wanted function is \(f(t)=(1/\mathrm e)\cdot \mathrm e^{\mathrm e^t}\), which can also be written as \(f(t)=\mathrm e^{-1+\mathrm e^t}\).

Exercise 4: A Homogeneous, Real System of Linear Differential Equations (By Hand)#

This exercise is intended to be solved purely by handed. A linear, real system of differential equations with constant coefficients is given as follows:

\[\begin{split} \left[\begin{array}{c} f_1'(t)\\ f_2'(t)\end{array}\right] = \left[\begin{array}{rr} 1 & 8\\ 1 &-1\end{array}\right] \cdot \left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right]. \end{split}\]

Question a#

Compute the eigenvalues of the coefficient matrix as well as their associated eigenspaces, and state based on this the general solution to the given differential equation system.

Answer

The eigenvalues are \(3\) and \(-3\). Their associated eigenspaces are

\[\begin{split}E_3=\mathrm{span}\left(\left[\begin{array}{c} 4\\ 1\end{array}\right]\right) \quad \text{ and } \quad E_{-3}=\mathrm{span}\left(\left[\begin{array}{c} -2\\ 1\end{array}\right]\right).\end{split}\]

According to Corollary 12.2.4 the general solution is hence

\[\begin{split}\left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right]=c_1 \cdot \left[\begin{array}{c} 4\\ 1\end{array}\right]\mathrm e^{3t}+c_2 \cdot \left[\begin{array}{c} -2\\ 1\end{array}\right]\mathrm e^{-3t}, \quad \text{ where } c_1,c_2 \in \mathbb{R}.\end{split}\]

Question b#

Determine the solution to the given system of differential equations that fulfills \(f_1(0)=0\) and \(f_2(0)=3\).

Hint

From Question a we know how the general solution looks. If \(t=0\) is inserted in the general solution, then we achieve a condition that the constants \(c_1\) and \(c_2\) must fulfill. Now try to solve for \(c_1\) and \(c_2\).

Answer

\[\begin{split}\left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right]=1 \cdot \left[\begin{array}{c} 4\\ 1\end{array}\right]\mathrm e^{3t}+2 \cdot \left[\begin{array}{c} -2\\ 1\end{array}\right]\mathrm e^{-3t}=\left[\begin{array}{c} 4\mathrm e^{3t}-4\mathrm e^{-3t}\\ \mathrm e^{3t}+2\mathrm e^{-3t}\end{array}\right].\end{split}\]

Exercise 5: A homogeneous, Real System of Linear Differential Equations (SymPy)#

We are given the following real system of differential equations:

\[\begin{split} \left\{ \begin{array}{rcl} f_1'(t) & = & 8f_1(t)+5f_2(t)\\ f_2'(t) & = & -10f_1(t)-7f_2(t). \end{array} \right. \end{split}\]

Question a#

Find a matrix \(\mathbf A\) and functions \(q_1(t)\) and \(q_2(t)\) such that

\[\begin{split} \left[\begin{array}{c} f_1'(t)\\ f_2'(t)\end{array}\right] = {\mathbf A} \cdot \left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right]+\left[\begin{array}{c} q_1(t)\\ q_2(t)\end{array}\right]. \end{split}\]

Is the system homogeneous or inhomogeneous?

Answer

It applies that

\[\begin{split} \left[\begin{array}{c} f_1'(t)\\ f_2'(t)\end{array}\right] = \left[ \begin{array}{cc} 8 & 5\\ -10 & -7\end{array} \right]\cdot \left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right]. \end{split}\]

Hence we have \(q_1(t)=0\), \(q_2(t)=0\), and

\[\begin{split}{\mathbf A}=\left[ \begin{array}{cc} 8 & 5\\ -10 & -7\end{array} \right].\end{split}\]

The system is homogeneous since both \(q_1(t)\) and \(q_2(t)\) are zero.

Question b#

Use SymPy to find the eigenvalues of \({\mathbf A}\) as well as bases for its corresponding eigenspaces. Use SymPy’s output to write out the general solution to the given real system of differential equations.

Hint

If the matrix is defined as A and the command A.eigenvects() is used, then we get the wanted ouput. Corollary 12.2.4 can now be used to find the wanted general solution.

Answer

The eigenvalues are \(-2\) and \(3\), while

\[\begin{split}E_{-2}=\mathrm{span}\left(\left[\begin{array}{c} -1/2\\ 1\end{array}\right]\right) \quad \text{and} \quad E_{3}=\mathrm{span}\left(\left[\begin{array}{c} -1\\ 1\end{array}\right]\right).\end{split}\]

We now achieve the wanted general solution from Corollary 12.2.4 in the textbook:

\[\begin{split} \left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right] = c_1\cdot \left[ \begin{array}{c} -1/2 \\ 1\end{array} \right]\cdot \mathrm e^{-2t}+c_2\cdot \left[ \begin{array}{c} -1 \\ 1\end{array} \right]\cdot \mathrm e^{3t}, \quad \text{ where } c_1,c_2 \in \mathbb{R}. \end{split}\]

Exercise 6: An Inhomogeneous, Real System of Linear Differential Equations#

We are given the following real system of differential equations:

\[\begin{split} \left\{ \begin{array}{rcl} f_1'(t) & = & 8f_1(t)+5f_2(t)+3\\ f_2'(t) & = & -10f_1(t)-7f_2(t)+1. \end{array} \right. \end{split}\]

Question a#

Check that the corresponding homogeneous system of differential equations is the system given in Exercise 5.

Question b#

We are informed that real numbers \(a\) and \(b\) exist such that the constant functions \(f_1(t)=a\) and \(f_2(t)=b\) constitute a particular solution to the given inhomogeneous system. Now compute \(a\) and \(b\). If you need inverse matrices you may compute them using SymPy.

Hint

Insert the functions \(f_1(t)=a\) and \(f_2(t)=b\) into the system. What must \(a\) and \(b\) fulfill?

Hint

If the matrix from Exercise 5 is denoted by \({\mathbf A}\), then we find that \(a\) and \(b\) must fulfill

\[\begin{split}\left[\begin{array}{c} 0\\ 0\end{array}\right]={\mathbf A} \cdot \left[\begin{array}{c} a\\ b\end{array}\right]+\left[\begin{array}{c} 3\\ 1\end{array}\right].\end{split}\]

Answer

\(a=-13/3\) and \(b=19/3\).

Question c#

Determine the general solution to the given inhomogeneous, real system of differential equations.

Hint

Definition 12.2.2 and your results from the previous questions can be used here.

Answer

\[\begin{split} \left[\begin{array}{c} f_1(t)\\ f_2(t)\end{array}\right] = \left[ \begin{array}{c} -13/3 \\ 19/3\end{array} \right]+c_1\cdot \left[ \begin{array}{c} -1/2 \\ 1\end{array} \right]\cdot \mathrm e^{-2t}+c_2\cdot \left[ \begin{array}{c} -1 \\ 1\end{array} \right]\cdot \mathrm e^{3t}, \quad \text{ where } c_1,c_2 \in \mathbb{R}. \end{split}\]

Exercise 7: Initial Conditions in a System of Linear Differential Equations#

We consider the same inhomogeneous, real system of differential equations as in Exercise 6. Compute the solution to the system that fulfills the initial conditions

\[\begin{split}\left[ \begin{array}{c} f_1(0) \\ f_2(0) \end{array} \right]=\left[ \begin{array}{c} 4 \\ -5\end{array} \right].\end{split}\]

You may compute inverse matrices using SymPy.

Answer

\[\begin{split}\left[ \begin{array}{c} f_1(t) \\ f_2(t) \end{array} \right]=\left[ \begin{array}{c} -13/3 \\ 19/3\end{array} \right]-6\cdot \left[ \begin{array}{c} -1/2 \\ 1\end{array} \right]\cdot \mathrm e^{-2t}-(16/3)\cdot \left[ \begin{array}{c} -1 \\ 1\end{array} \right]\cdot \mathrm e^{3t}=\left[ \begin{array}{c} -13/3+3 \mathrm e^{-2t}+(16/3)\mathrm e^{3t} \\ 19/3-6 \mathrm e^{-2t}-(16/3)\mathrm e^{3t}\end{array} \right].\end{split}\]

Exercise 8: The Solution Formula for Inhomogeneous, 1st-Order Differential Equations#

We are given the real differential equation \(f'(t)+f(t)/t=3t.\) We will assume that \(t>0\).

Question a#

Compute the general solution to the differential equation.

Hint

The differential equation can be rewritten to \(f'(t)=\frac{-1}{t} \cdot f(t)+3t\). Hence Theorem 12.1.1 from the textbook applies.

Hint

The previous hint implies that Theorem 12.1.1 can be used with \(a(t)=-1/t\) and \(q(t)=3t\). Example 12.1.2 in the textbook contains a walk-through where \(a(t)=-1/t\) just like here.

Answer

\(f(t)=t^2+c/t\), where \(c \in \mathbb{R}\).

Question b#

Compute the particular solution to the differential equation that fulfills the initial condition \(f(1)=5\).

Answer

\(f(t)=t^2+4/t\).

Exercise 9: A Tricky Coefficient Matrix#

Let \(\lambda\) be a real number, and consider the following real differential equation system:

\[\begin{split} \left[\begin{array}{c} f_1'(t)\\ f_2'(t)\\ f_3'(t)\end{array}\right] = \left[\begin{array}{rrr} \lambda & 1 & 0\\ 0 & \lambda &1\\ 0 & 0 & \lambda\end{array}\right] \cdot \left[\begin{array}{c} f_1(t)\\ f_2(t)\\ f_3(t)\end{array}\right]. \end{split}\]

What is the general solution of the system?

Hint

The usual methods do not work since the coefficient matrix has one eigenvalue \(\lambda\) that has an algebraic multiplicity of \(3\) and a geometric multiplicity of \(1\). Instead, try to get inspiration from Example 12.2.7 in the textbook.

Hint

That the following two vectors are solutions can be shown in a way similar to that in Example 12.2.7:

\[\begin{split}\left[ \begin{array}{c} f_1(t) \\ f_2(t) \\ f_3(t) \end{array} \right]=\left[ \begin{array}{c} \mathrm e^{\lambda t}\\ 0 \\ 0\end{array} \right] \quad \text{ and } \left[ \begin{array}{c} f_1(t) \\ f_2(t) \\ f_3(t) \end{array} \right] = \left[ \begin{array}{c} t\mathrm e^{\lambda t} \\ \mathrm e^{\lambda t} \\ 0\end{array} \right].\end{split}\]

We are now just missing a solution that is linearly independent from these two in order to find the general solution.

Hint

Try finding a solution in the form

\[\begin{split}\left[ \begin{array}{c} f_1(t) \\ f_2(t) \\ f_3(t) \end{array} \right]= \left[ \begin{array}{c} at^2\mathrm e^{\lambda t} \\ t\mathrm e^{\lambda t} \\ \mathrm e^{\lambda t}\end{array} \right],\end{split}\]

where \(a \in \mathbb{R}.\)

Answer

\[\begin{split}\left[ \begin{array}{c} f_1(t) \\ f_2(t) \\ f_3(t) \end{array} \right]=c_1\cdot \left[ \begin{array}{c} \mathrm e^{\lambda t}\\ 0 \\ 0\end{array} \right]+c_2 \cdot \left[ \begin{array}{c} t\mathrm e^{\lambda t} \\ \mathrm e^{\lambda t} \\ 0\end{array} \right]+c_3 \cdot \left[ \begin{array}{c} (1/2)t^2 \mathrm e^{\lambda t} \\ t\mathrm e^{\lambda t}\\ \mathrm e^{\lambda t}\end{array} \right], \quad \text{ where } c_1,c_2,c_3 \in \mathbb{R}.\end{split}\]