Systems of ODEs Practice

Write the state vector as

Then the system is

The system is linear because it has the form

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

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

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

arbitrary constants.

An equilibrium occurs when both derivatives are zero.

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

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

Combine the choices to get the equilibria:

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

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

The mode has the factor

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

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.

The trace is the sum of the diagonal entries:

The determinant is

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

That makes the origin an asymptotically stable node.

The linearization is

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

Compute

So the characteristic polynomial is

The eigenvalues are

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

The first component is

so

The second component is

so

Thus

Each eigenpair gives an eigenmode solution.

So the general solution is

The characteristic polynomial is

The discriminant is

so the eigenvalues are complex.

Solve the quadratic:

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

Since $u=x$, we have

Also,

So the first-order system is

Compute the partial derivatives:

So

A particular solution can be written as

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

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

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

Set both derivatives equal to zero.

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

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

Check the intersections:

These are the equilibria.

Since $u=x$,

Also,

so

Thus the first-order system is

At equilibrium, set both derivatives equal to zero:

From the second equation,

so

Substitute into the first equation:

Combine terms:

So

Then

The equilibrium point is

The coefficient matrix is

Its trace and determinant are

The discriminant is

so the eigenvalues are complex.

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

First compute the Jacobian:

At $(0,0)$,

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.

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

The system can be written as

Then

So the general solution is

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

Solving gives

Now shift variables:

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

The coefficient matrix is

whose eigenvalues are

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

Set both derivatives equal to zero:

This gives the equilibria

Now compute the Jacobian:

At $(0,0)$,

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

At $(2,0)$,

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

At $(2,1)$,

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

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

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

Since

we get

So the first-order system is

The coefficient matrix is

Its characteristic polynomial is

So the eigenvalues are

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