Let's analyze the given autonomous system of differential equations:

$\frac{dx}{dt} = x,$ $\frac{dy}{dt} = x^2 + y.$

Part (a): Find the Critical Points

Critical points occur where both $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} = 0$.

1. For $\frac{dx}{dt} = 0$:

$x = 0.$

2. For $\frac{dy}{dt} = 0$:

$x^2 + y = 0.$

Substituting $x = 0$ into $x^2 + y = 0$:

$0 + y = 0 \implies y = 0.$

So, the only critical point is at $(0,0)$.

Part (b): Solve the Phase Plane Equation and Classify the Critical Points

To classify the critical points, we linearize the system around the critical point $(0,0)$.

1. Write the system as a vector function:

$\mathbf{F}(x, y) = \begin{pmatrix} x \\ x^2 + y \end{pmatrix}.$

2. Compute the Jacobian matrix $J$ at the critical point $(0,0)$:

$J = \begin{pmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 2x & 1 \end{pmatrix}.$

Evaluating at $(0,0)$:

$J(0,0) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$

3. The eigenvalues of $J(0,0)$ are the solutions to $\det(J - \lambda I) = 0$:

$\det \begin{pmatrix} 1 - \lambda & 0 \\ 0 & 1 - \lambda \end{pmatrix} = (1 - \lambda)^2 = 0.$

Thus, $\lambda_1 = 1$ and $\lambda_2 = 1$.

Since both eigenvalues are positive, the critical point $(0,0)$ is an unstable node.

Part (c): Find the General Solution to the System and Determine its Stability

To find the general solution, solve the individual differential equations.

1. For $\frac{dx}{dt} = x$:

$\frac{dx}{x} = dt \implies \int \frac{1}{x} \, dx = \int 1 \, dt \implies \ln|x| = t + C_1 \implies x = C_2 e^t,$

where $C_2 = e^{C_1}$ is a constant.

2. Substitute $x = C_2 e^t$ into $\frac{dy}{dt} = x^2 + y$:

$\frac{dy}{dt} = (C_2 e^t)^2 + y \implies \frac{dy}{dt} - y = C_2^2 e^{2t}.$

This is a first-order linear non-homogeneous differential equation. The integrating factor $\mu(t)$ is:

$\mu(t) = e^{-\int 1 \, dt} = e^{-t}.$

Multiply through by the integrating factor:

$e^{-t} \frac{dy}{dt} - e^{-t} y = C_2^2 e^t.$

The left side is the derivative of $y e^{-t}$:

$\frac{d}{dt} (y e^{-t}) = C_2^2 e^t.$

Integrate both sides with respect to $t$:

$y e^{-t} = C_2^2 \int e^t \, dt = C_2^2 e^t + C_3.$

So,

$y = C_2^2 e^{2t} + C_3 e^t.$

Combining the solutions for $x$ and $y$:

$x(t) = C_2 e^t,$ $y(t) = C_2^2 e^{2t} + C_3 e^t.$

Stability and Integral Curves

• The general solution shows that as $t \to \infty$, both $x(t)$ and $y(t)$ grow exponentially.
• This confirms that the critical point $(0,0)$ is unstable.
• Integral curves in the phase plane will diverge away from the origin as time progresses.

Would you like details or have any questions?

Here are five more questions you might want to ask:

1. How do you solve non-linear differential equations?
2. What are the methods to determine the stability of a system of differential equations?
3. Can you explain the concept of the phase plane and how to sketch it for a given system?
4. How do you apply the Lyapunov method for stability analysis?
5. What are the differences between nodes, saddles, and spirals in the classification of critical points?

Tip: When analyzing systems of differential equations, always consider both the eigenvalues and the eigenvectors of the Jacobian matrix to fully understand the behavior near critical points.