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 \(f(t)=c\cdot \mathrm e^{(\mathrm e^t)},\) we should try to find 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#

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 13.2.4 in the textbook 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: Another homogeneous, Real System of Linear Differential Equations#

We are given the following real system of differential equations:

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

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

We see 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#

We are being informed that the matrix

\[\begin{split}\mathbf A=\begin{bmatrix}8&5\\-10&-7\end{bmatrix}\end{split}\]

has eigenvalues \(-2\) and \(3\) with eigenspaces

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

Someone is now proposing that the general solution to the given real system of differential equations is

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

Is this proposed general solution correct?

Hint

By using Corollary 13.2.4 with the given information we arrive at a general solution looking as follows:

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

Does this general solution express other solutions than the proposed general solution does when \(c_1\) and \(c_2\) are chosen freely from \(\Bbb R\)?

Answer

Yes, the proposed general solution is correct. And the alternative general solution mentioned in the hint is also correct, because they are equivalent and express the same solution set.

One way to see that is to notice that \(\begin{bmatrix} 1\\-2 \end{bmatrix}=-2\cdot \begin{bmatrix} -1/2\\1 \end{bmatrix}\) and \(\begin{bmatrix} 1\\-1 \end{bmatrix}=-1\cdot \begin{bmatrix} -1\\1 \end{bmatrix}\), which reveals that \(\begin{bmatrix} 1\\-2 \end{bmatrix}\) and \(\begin{bmatrix} 1\\-1 \end{bmatrix}\) are eigenvectors of \(\mathbf A\) corresponding to the eigenvalues \(-2\) and \(3\), respectively. Hence

\[\begin{split}\left(\begin{bmatrix} 1\\-2 \end{bmatrix},\begin{bmatrix} 1\\-1 \end{bmatrix}\right)\end{split}\]

is an ordered basis for \(\Bbb R^2\) consisting of eigenvectors of the matrix \(\mathbf A\), and Corollary 13.2.4 then tells that the proposed general solution indeed is correct.

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. Find these \(a\) and \(b\).

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

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

The easiest way to determine \(\mathbf A^{-1}\) is to use the formula from Example 8.1.3 in the textbook.

Answer

\(a=-13/3\) and \(b=19/3\). Based on the previous hint we have for example that

\[\begin{split}\begin{bmatrix} a\\b \end{bmatrix}=\mathbf A^{-1}\begin{bmatrix} -3\\-1 \end{bmatrix}=\frac1{-6}\begin{bmatrix} -7&-5\\10&8 \end{bmatrix}\begin{bmatrix} -3\\-1 \end{bmatrix}=\frac16\begin{bmatrix} -7&-5\\10&8 \end{bmatrix}\begin{bmatrix} 3\\1 \end{bmatrix}=\frac16\begin{bmatrix} -26\\38 \end{bmatrix}\end{split}\]

Question c#

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

Hint

Definition 13.2.2 and your results from the previous Questions can be used here.

Answer

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

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

\[\begin{split}\left[ \begin{array}{c} f_1(0) \\ f_2(0) \end{array} \right]=\left[ \begin{array}{c} 2/3 \\ 1/3\end{array} \right].\end{split}\]

Answer

\[\begin{split}\left[ \begin{array}{c} f_1(t) \\ f_2(t) \end{array} \right]=\left[ \begin{array}{c} -13/3 \\ 19/3\end{array} \right]+1\cdot \left[ \begin{array}{c} 1 \\ -2\end{array} \right]\cdot \mathrm e^{-2t}+4\cdot \left[ \begin{array}{c} 1 \\ -1\end{array} \right]\cdot \mathrm e^{3t}=\left[ \begin{array}{c} -13/3+\mathrm e^{-2t}+4\mathrm e^{3t} \\ 19/3-2 \mathrm e^{-2t}-4\mathrm e^{3t}\end{array} \right].\end{split}\]

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 13.1.1 from the textbook applies.

Hint

The previous hint implies that Theorem 13.1.1 can be used with \(a(t)=-1/t\) and \(q(t)=3t\). Example 13.1.2 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 13.2.7 in the textbook.

Hint

That the following two vectors are solutions can be shown in a way similar to that in Example 13.2.7:

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

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

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