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Give step-by-step solution with explanation and final answer:Exercise 1.6.3: Consider x" = x2. a) Draw the phase diagram, find the critical points, and mark them stable or unstable. b) Sketch typical solutions of the equation. ¢) Find Jim x(t) for the solution with the initial condition x(0) = —1.

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Give step-by-step solution with explanation and final answer:Uploaded ImageExercise 1.6.3: Consider x" = x2. a) Draw the phase diagram, find the critical points, and mark them stable or unstable. b) Sketch typical solutions of the equation. ¢) Find Jim x(t) for the solution with the initial condition x(0) = —1.

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Asked by: gvr
Created at: 2025-06-18 10:51:08

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Let's solve each part step-by-step: ## Given: \[ x' = x^2 \] --- ## (a) **Phase diagram, Critical points, Stability** ### **Critical points** Critical points occur where \( x' = 0 \): \[ x^2 = 0 \implies x = 0 \] So, the only critical point is **\( x = 0 \)**. ### **Stability** - For \( x > 0 \): \( x^2 > 0 \) so \( x' > 0 \) — solution moves away from 0 (increases). - For \( x < 0 \): \( x^2 > 0 \) so \( x' > 0 \) — solution moves away from 0 (increases toward 0, but remains negative). Thus, the critical point \( x = 0 \) is **unstable**. ### **Phase diagram** - Draw a line for the \( x \)-axis. - Mark the critical point at \( x = 0 \). - Draw arrows pointing **right** (→) on both sides of \( x = 0 \) (since \( x' > 0 \) everywhere except at \( x = 0 \)). --- ## (b) **Typical Solutions** Let's solve \( x' = x^2 \): \[ \frac{dx}{dt} = x^2 \implies \frac{dx}{x^2} = dt \] Integrate both sides: \[ \int x^{-2} dx = \int dt \implies -x^{-1} = t + C \implies -\frac{1}{x} = t + C \implies x = -\frac{1}{t + C} \] **Typical solutions:** - For \( x(0) > 0 \): \( x(t) \) increases and blows up to \( +\infty \) in finite time. - For \( x(0) < 0 \): \( x(t) \) increases (becomes less negative), but stays negative and approaches 0 as \( t \to \infty \). - For \( x(0) = 0 \): \( x(t) = 0 \) (constant solution). --- ## (c) **Limit as \( t \to \infty \) with \( x(0) = -1 \)** From above, the solution is: \[ x(t) = -\frac{1}{t + C} \] Find \( C \) using \( x(0) = -1 \): \[ x(0) = -1 \implies -1 = -\frac{1}{0 + C} \implies 1 = \frac{1}{C} \implies C = 1 \] So, \[ x(t) = -\frac{1}{t + 1} \] Now, compute the limit: \[ \lim_{t \to \infty} x(t) = \lim_{t \to \infty} -\frac{1}{t + 1} = 0 \] --- ## **Summary of Answers** ### **(a)** - Critical point: \( x = 0 \) (unstable). - Phase line: arrows pointing right everywhere except at \( x = 0 \). ### **(b)** - General solution: \( x(t) = -\frac{1}{t + C} \) - Solutions with \( x(0) > 0 \) blow up in finite time. - Solutions with \( x(0) < 0 \) approach 0 from below as \( t \to \infty \). ### **(c)** \[ \boxed{0} \] That is, \[ \lim_{t \to \infty} x(t) = 0 \quad \text{for} \quad x(0) = -1. \]

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