Let's solve each part step by step. --- ### **Given:** \[ x' = x^2 \] --- ## **a) Phase diagram, critical points, stability** ### **Step 1: Find the critical points** Critical points occur where \( x' = 0 \): \[ x^2 = 0 \implies x = 0 \] So, the only critical point is at \( x = 0 \). --- ### **Step 2: Analyze stability** - For \( x > 0 \): \( x^2 > 0 \implies x' > 0 \), so \( x \) increases. - For \( x < 0 \): \( x^2 > 0 \implies x' > 0 \), so \( x \) increases (toward 0). #### **Phase Line:** - Draw a line with \( x \) values. - At \( x = 0 \), mark a point (critical point). - For \( x > 0 \), draw arrows pointing **away from 0** (to the right). - For \( x < 0 \), draw arrows pointing **toward 0** (to the right). #### **Stability:** - At \( x = 0 \), both sides point **away from 0** (for \( x>0 \)) and **toward 0** (for \( x<0 \)), but notice for negative values, \( x \) increases toward 0, but as soon as it crosses to positive, it moves away rapidly (since \( x^2 \) grows fast). - **\( x = 0 \) is an unstable critical point** (since any positive perturbation causes \( x \) to blow up). --- ## **b) Sketch typical solutions** ### **Step 1: Solve the equation analytically** \[ \frac{dx}{dt} = x^2 \] Separate variables: \[ \frac{dx}{x^2} = dt \] Integrate both sides: \[ \int x^{-2} dx = \int dt \] \[ - x^{-1} = t + C \] \[ \frac{1}{x} = - (t + C) \] \[ x(t) = \frac{1}{- (t + C)} \] ### **Step 2: Typical solutions** - For \( x(0) = x_0 \): \[ x(0) = \frac{1}{-C} \implies C = -\frac{1}{x_0} \] \[ x(t) = \frac{1}{- (t - \frac{1}{x_0})} = \frac{x_0}{1 - x_0 t} \] - **If \( x_0 > 0 \):** denominator hits zero at \( t = \frac{1}{x_0} \) (blows up in finite time). - **If \( x_0 < 0 \):** denominator never hits zero for positive \( t \), approaches zero from below as \( t \to \infty \). **Sketch:** - All solutions with \( x_0 > 0 \): blow up in finite time. - Solutions with \( x_0 < 0 \): approach 0 from below as \( t \to \infty \). - \( x_0 = 0 \): stays at 0 (critical point). --- ## **c) Limit for \( x(0) = -1 \) as \( t \to \infty \)** From above: \[ x(t) = \frac{-1}{1 + t} \] As \( t \to \infty \), \( x(t) \to 0^- \). ### **Final Answer:** \[ \boxed{0} \] --- ### **Summary Table** | Part | Result | |------|--------| | (a) | Critical point at \( x=0 \), unstable. | | (b) | Solutions: \( x(t) = \frac{x_0}{1 - x_0 t} \). Positive initial values blow up; negative initial values approach 0 from below. | | (c) | \( \lim_{t\to\infty} x(t) = 0 \) for \( x(0) = -1 \). | If you need the phase diagram or a sketch, let me know!