Systems of ODEs Practice

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1.1Write a System in Matrix Form

Exam I | Problem 1.1 |

Rewrite the system in matrix form:

\begin{aligned} x' &= 2x - y \\ y' &= 4x + 3y \end{aligned}

Solution

Write the state vector as

\mathbf{x} = \begin{bmatrix} x \\ y \end{bmatrix}.

Then the system is

\mathbf{x}' = \begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix} \mathbf{x}.

1.2Classify a Linear System

Exam I | Problem 1.2 |

Classify the system as linear, autonomous, homogeneous, or nonhomogeneous:

\mathbf{x}' = \begin{bmatrix} 1 & 0 \\ 2 & -3 \end{bmatrix} \mathbf{x} \begin{bmatrix} \sin t \\ 1 \end{bmatrix}.

Solution

The system is linear because it has the form

\mathbf{x}' = A\mathbf{x} + \mathbf{g}(t).

It is nonautonomous because the forcing term depends on $t$.

It is nonhomogeneous because $\mathbf{g}(t) \neq \mathbf{0}$.

1.3Count the Constants in a 4x4 System

Exam I | Problem 1.3 |

A first-order linear homogeneous system has size $4 \times 4$.

How many arbitrary constants appear in the general solution?

Solution

A first-order linear homogeneous system of size $n \times n$ has $n$ independent solution constants.

Here $n = 4$, so the general solution depends on

$$ 4 $$

arbitrary constants.

1.4Find All Equilibria of a Nonlinear System

Exam I | Problem 1.4 |

Find all equilibria of the system

\begin{aligned} x' &= x(3-x) \\ y' &= y(y-2) \end{aligned}

Solution

An equilibrium occurs when both derivatives are zero.

From $x' = x(3-x) = 0$, we get

x = 0 \quad \text{or} \quad x = 3.

From $y' = y(y-2) = 0$, we get

y = 0 \quad \text{or} \quad y = 2.

Combine the choices to get the equilibria:

(0,0),\ (0,2),\ (3,0),\ (3,2).

1.5Use an Eigenpair to Write a Solution

Exam I | Problem 1.5 |

Suppose $A\mathbf{v} = -2\mathbf{v}$.

What solution of $\mathbf{x}' = A\mathbf{x}$ does this eigenpair generate?

Solution

An eigenpair $(\lambda,\mathbf{v})$ gives a solution of the form

\mathbf{x}(t) = e^{\lambda t}\mathbf{v}.

Here $\lambda = -2$, so the solution is

\mathbf{x}(t) = e^{-2t}\mathbf{v}.

1.6Interpret a Negative Eigenvalue

Exam I | Problem 1.6 |

A mode of a system has eigenvalue $\lambda = -5$.

What does that tell you about the behavior of that mode?

Solution

The mode has the factor

e^{-5t}.

As $t$ increases, $e^{-5t} \to 0$, so that mode decays exponentially.

1.7Interpret Complex Eigenvalues

Exam I | Problem 1.7 |

A system has eigenvalues

\lambda = 1 \pm 4i.

What qualitative behavior should you expect?

Solution

Complex eigenvalues produce oscillation, and the real part controls growth or decay.

Here the real part is $1$, so the oscillations are multiplied by $e^t$.

That means the motion oscillates while growing in size, so trajectories spiral outward.

1.8Compute Trace and Determinant

Exam I | Problem 1.8 |

For the matrix

A = \begin{bmatrix} 3 & -2 \\ 5 & -1 \end{bmatrix},

find $\operatorname{tr}(A)$ and $\det(A)$.

Solution

The trace is the sum of the diagonal entries:

\operatorname{tr}(A) = 3 + (-1) = 2.

The determinant is

\det(A) = 3(-1) - (-2)(5) = -3 + 10 = 7.

1.9Classify Stability from Eigenvalues

Exam I | Problem 1.9 |

A linear $2 \times 2$ system has eigenvalues $-1$ and $-4$.

What type of equilibrium does the origin have?

Solution

Both eigenvalues are real and negative, so all nearby solutions decay toward the origin.

That makes the origin an asymptotically stable node.

1.10State the Linearization Formula

Exam I | Problem 1.10 |

Near an equilibrium point $\mathbf{x}^*$, what first-order approximation do you use for a nonlinear system?

Solution

The linearization is

\mathbf{x}' \approx J(\mathbf{x}^*)(\mathbf{x} - \mathbf{x}^*),

where $J(\mathbf{x}^*)$ is the Jacobian matrix evaluated at the equilibrium.

Difficulty

Integrated

2.1Find a Characteristic Polynomial

Exam II | Problem 2.1 |

For

A = \begin{bmatrix} 1 & 2 \\ 0 & 4 \end{bmatrix},

find the characteristic polynomial and the eigenvalues.

Solution

Compute

\det(A - \lambda I) = \det\begin{bmatrix} 1-\lambda & 2 \\ 0 & 4-\lambda \end{bmatrix} =(1-\lambda)(4-\lambda).

So the characteristic polynomial is

\lambda^2 - 5\lambda + 4.

The eigenvalues are

\lambda = 1,\ 4.

2.2Solve a Diagonal System

Exam II | Problem 2.2 |

Solve the initial value problem

\mathbf{x}' = \begin{bmatrix} -2 & 0 \\ 0 & 3 \end{bmatrix} \mathbf{x}, \qquad \mathbf{x}(0) = \begin{bmatrix} 4 \\ -1 \end{bmatrix}.

Solution

Because the matrix is diagonal, each component solves its own scalar equation.

The first component is

x_1' = -2x_1, \qquad x_1(0)=4,

x_1(t) = 4e^{-2t}.

The second component is

x_2' = 3x_2, \qquad x_2(0)=-1,

x_2(t) = -e^{3t}.

Thus

\mathbf{x}(t) = \begin{bmatrix} 4e^{-2t} \\ -e^{3t} \end{bmatrix}.

2.3Build a General Solution from Eigenpairs

Exam II | Problem 2.3 |

Suppose a matrix $A$ has eigenpairs

(2, \begin{bmatrix}1 \\ 1\end{bmatrix}) \quad \text{and} \quad (-1, \begin{bmatrix}1 \\ -1\end{bmatrix}).

Write the general solution of $\mathbf{x}' = A\mathbf{x}$.

Solution

Each eigenpair gives an eigenmode solution.

So the general solution is

\mathbf{x}(t) = c_1 e^{2t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2 e^{-t} \begin{bmatrix} 1 \\ -1 \end{bmatrix}.

2.4Classify from Trace and Determinant

Exam II | Problem 2.4 |

A $2 \times 2$ matrix has trace $-2$ and determinant $5$.

Determine whether the eigenvalues are real or complex, and classify the equilibrium.

Solution

The characteristic polynomial is

\lambda^2 - (\operatorname{tr} A)\lambda + \det(A) = \lambda^2 + 2\lambda + 5.

The discriminant is

$$ 2^2 - 4(5) = -16, $$

so the eigenvalues are complex.

Solve the quadratic:

\lambda = \frac{-2 \pm \sqrt{-16}}{2} = -1 \pm 2i.

Because the real part is negative, the equilibrium is a stable spiral.

2.5Convert a Mass-Spring Equation

Exam II | Problem 2.5 |

A spring-mass system satisfies

$$ x'' + 2x' + 9x = 0. $$

Let $u=x$ and $v=x'$. Write the equivalent first-order system.

Solution

Since $u=x$, we have

$$ u' = x' = v. $$

Also,

$$ v' = x'' = -2x' - 9x = -2v - 9u. $$

So the first-order system is

\begin{aligned} u' &= v \\ v' &= -9u - 2v. \end{aligned}

2.6Find the Jacobian Matrix

Exam II | Problem 2.6 |

Find the Jacobian matrix of

\mathbf{f}(x,y) = \begin{bmatrix} x^2 + y \\ xy - 1 \end{bmatrix}.

Solution

Compute the partial derivatives:

\frac{\partial f_1}{\partial x} = 2x, \qquad \frac{\partial f_1}{\partial y} = 1,

\frac{\partial f_2}{\partial x} = y, \qquad \frac{\partial f_2}{\partial y} = x.

J(x,y) = \begin{bmatrix} 2x & 1 \\ y & x \end{bmatrix}.

2.7Write the Variation-of-Parameters Formula

Exam II | Problem 2.7 |

For the forced system

\mathbf{x}' = A\mathbf{x} + \mathbf{g}(t),

let $\Phi(t)$ be a fundamental matrix for $\mathbf{x}' = A\mathbf{x}$.

What formula gives a particular solution?

Solution

A particular solution can be written as

\mathbf{x}_p(t) = \Phi(t)\int \Phi(t)^{-1}\mathbf{g}(t)\,dt.

The full solution is then the sum of the homogeneous and particular parts.

2.8Handle a Repeated Eigenvalue

Exam II | Problem 2.8 |

If a $2 \times 2$ system has a repeated eigenvalue $\lambda$ and only one eigenvector $\mathbf{v}$, what is a common form of a second independent solution?

Solution

In the defective case, a second solution often has the form

e^{\lambda t}(\mathbf{v} + t\mathbf{w}),

where $\mathbf{w}$ is a generalized eigenvector.

Difficulty

Applied

3.1Find Equilibria in a Predator-Prey Model

Final | Problem 3.1 |

A predator-prey model is

\begin{aligned} x' &= x(3-y) \\ y' &= y(x-2) \end{aligned}

where $x$ and $y$ are the population levels.

Find all equilibria.

Solution

Set both derivatives equal to zero.

From $x' = x(3-y)=0$, we get

x=0 \quad \text{or} \quad y=3.

From $y' = y(x-2)=0$, we get

y=0 \quad \text{or} \quad x=2.

Check the intersections:

(0,0), \quad (2,0), \quad (2,3).

These are the equilibria.

3.2Convert a Damped Oscillator

Final | Problem 3.2 |

A mass-spring-damper system satisfies

$$ 2x'' + 6x' + 8x = 0. $$

Let $u=x$ and $v=x'$.

Write the equivalent first-order system.

Solution

Since $u=x$,

$$ u' = v. $$

Also,

$$ 2v' + 6v + 8u = 0, $$

$$ v' = -3v - 4u. $$

Thus the first-order system is

\begin{aligned} u' &= v \\ v' &= -4u - 3v. \end{aligned}

3.3Find the Steady State of a Forced System

Final | Problem 3.3 |

A model is given by

\begin{aligned} x' &= -3x + y + 4 \\ y' &= 2x - 2y + 1 \end{aligned}

Find the equilibrium point.

Solution

At equilibrium, set both derivatives equal to zero:

$$ -3x + y + 4 = 0, $$

$$ 2x - 2y + 1 = 0. $$

From the second equation,

x - y = -\frac{1}{2},

y = x + \frac{1}{2}.

Substitute into the first equation:

-3x + \left(x + \frac{1}{2}\right) + 4 = 0.

Combine terms:

-2x + \frac{9}{2} = 0.

x = \frac{9}{4}.

Then

y = \frac{9}{4} + \frac{1}{2} = \frac{11}{4}.

The equilibrium point is

\left(\frac{9}{4}, \frac{11}{4}\right).

3.4Classify a Circuit Model

Final | Problem 3.4 |

A circuit model is

\begin{aligned} x' &= -x + 2y \\ y' &= -3x - 4y \end{aligned}

Determine the type of equilibrium at the origin.

Solution

The coefficient matrix is

A = \begin{bmatrix} -1 & 2 \\ -3 & -4 \end{bmatrix}.

Its trace and determinant are

\operatorname{tr}(A) = -5,

\det(A) = (-1)(-4) - (2)(-3) = 10.

The discriminant is

$$ (-5)^2 - 4(10) = 25 - 40 = -15, $$

so the eigenvalues are complex.

Because the trace is negative, the real part is negative, so the origin is a stable spiral.

3.5Linearize a Nonlinear System at the Origin

Final | Problem 3.5 |

Consider the nonlinear system

\begin{aligned} x' &= x(1-y) \\ y' &= y(2-x) \end{aligned}

Use the linearization at $(0,0)$ to determine the local behavior of the origin.

Solution

First compute the Jacobian:

J(x,y) = \begin{bmatrix} 1-y & -x \\ -y & 2-x \end{bmatrix}.

At $(0,0)$,

J(0,0) = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}.

The eigenvalues are $1$ and $2$, both positive.

So the origin is an unstable node for the linearized system, and the nonlinear system is locally unstable there.

Difficulty

Challenge

4.1Solve a Defective Linear System

Final | Problem 4.1 |

Solve the system

\mathbf{x}' = \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix} \mathbf{x}.

Solution

The matrix has repeated eigenvalue $3$ and only one eigenvector, so it is defective.

The system can be written as

A = 3I + N, \qquad N = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}.

Then

e^{At} = e^{3t}e^{Nt} = e^{3t} \begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix}.

So the general solution is

\mathbf{x}(t) = c_1 e^{3t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 e^{3t} \begin{bmatrix} t \\ 1 \end{bmatrix}.

4.2Shift a Forced System and Classify It

Final | Problem 4.2 |

Consider the forced system

\begin{aligned} x' &= 2x - y + 1 \\ y' &= x + 2y - 3 \end{aligned}

Find the equilibrium point, shift variables to move the equilibrium to the origin, and classify the shifted linear system.

Solution

First find the equilibrium by setting both derivatives equal to zero:

$$ 2x - y + 1 = 0, $$

$$ x + 2y - 3 = 0. $$

Solving gives

\left(\frac{1}{5}, \frac{7}{5}\right).

Now shift variables:

u = x - \frac{1}{5}, \qquad v = y - \frac{7}{5}.

Because the constant terms disappear after shifting, the new system is

\begin{aligned} u' &= 2u - v \\ v' &= u + 2v. \end{aligned}

The coefficient matrix is

\begin{bmatrix} 2 & -1 \\ 1 & 2 \end{bmatrix},

whose eigenvalues are

2 \pm i.

The real part is positive, so the equilibrium is an unstable spiral.

4.3Analyze a Nonlinear System with Linearization

Final | Problem 4.3 |

Consider

\begin{aligned} x' &= x(1-y) \\ y' &= y(x-2) \end{aligned}

Find the equilibria and use linearization to determine which ones you can classify directly.

Solution

Set both derivatives equal to zero:

x(1-y)=0, \qquad y(x-2)=0.

This gives the equilibria

(0,0), \quad (2,0), \quad (2,1).

Now compute the Jacobian:

J(x,y) = \begin{bmatrix} 1-y & -x \\ y & x-2 \end{bmatrix}.

At $(0,0)$,

J(0,0) = \begin{bmatrix} 1 & 0 \\ 0 & -2 \end{bmatrix}.

The eigenvalues are $1$ and $-2$, so $(0,0)$ is a saddle and is unstable.

At $(2,0)$,

J(2,0) = \begin{bmatrix} 1 & -2 \\ 0 & 0 \end{bmatrix}.

One eigenvalue is $0$, so linearization is inconclusive there.

At $(2,1)$,

J(2,1) = \begin{bmatrix} 0 & -2 \\ 1 & 0 \end{bmatrix},

whose eigenvalues are purely imaginary, so linearization is also inconclusive there.

Only $(0,0)$ can be classified directly from the linearization.

4.4Classify a Damped Oscillator from Its System

Final | Problem 4.4 |

A damped oscillator satisfies

$$ x'' + 4x' + 13x = 0. $$

Let $u=x$ and $v=x'$. Convert the equation to a first-order system, find the eigenvalues of the coefficient matrix, and classify the motion.

Solution

With $u=x$ and $v=x'$, we have

$$ u' = v. $$

Since

$$ x'' + 4x' + 13x = 0, $$

we get

$$ v' = -4v - 13u. $$

So the first-order system is

\begin{aligned} u' &= v \\ v' &= -13u - 4v. \end{aligned}

The coefficient matrix is

A = \begin{bmatrix} 0 & 1 \\ -13 & -4 \end{bmatrix}.

Its characteristic polynomial is

\det(A - \lambda I) = \lambda^2 + 4\lambda + 13.

So the eigenvalues are

\lambda = -2 \pm 3i.

Because the real part is negative and the eigenvalues are complex, the motion is a stable spiral.