do b and c **Exercise 2.3.3**For each Ordinary Differential Equation (ODE), sketch a phase portrait by hand, following the procedure of Examples 2.10, 2.11, and 2.12. Classify each fixed point as asymptotically stable or unstable. Use the result to sketch solutions for the given initial conditions on a pair of $u$ axes with a reasonable range for $u$.(a) $ u'(v) = -u(v) $, sketch solutions with $ u(0) = 2 $ and $ u(0) = -2 $.(b) $ v'(u) = 11 - 2v(t) $, sketch solutions with $ v(0) = 0 $ and $ v(0) = 15 $.(c) $ u'(v) = -ku(v) $ (a positive constant), sketch solutions with $ v(0) = 0 $ and $ v(0) = 15/k $.(d) $ u'(v) = -(u(t) - 1)(u(t) - 3) $, sketch solutions with $ u(0) = 1/2 $, $ u(0) = 2 $, and $ u(0) = 4 $.(e) $ u'(t) = r(u(t)(1 - u(t)) - u(t)/10 $ (the harvested logistic equation (1.12) with $ r = 1 $, $ K = 1 $, and $ h = 1/10 $), sketch solutions with $ u(0) = 1/2 $ and $ u(0) = 3/2 $. Note only $ u \geq 0 $ makes physical sense here. What is the long-term fate of the species?(f) $ u'(t) = u(t)(1 - \frac{u(t)}{2}) - 2u(t) $ (the harvested logistic equation (1.12) with $ r = 1 $, $ K = 1 $, and $ h = 2 $), sketch solutions with $ u(0) = 1/2 $ and $ u(0) = 3/2 $. Note only $ u \geq 0 $ makes physical sense here. What is the long-term fate of **Example 2.10** Let us sketch a phase portrait for the autonomous ODE $ u'(t) = u^2(t) - 1 $. In this case, the ODE is $ u' = f(u) $ with $ f(u) = u^2 - 1 $ and the fixed points are the solutions to $ f(u) = 0 $, that is, $ u^2 - 1 = 0 $. These fixed points are $ u = -1 $ and $ u = 1 $, shown as the blue dots on the $ u $-axis in Figure 2.7.The fixed points divide the $ u $-axis into intervals $(-\infty, -1)$, $(-1, 1)$, and $(1, \infty)$. If $ u < -1 $ then $ f(u) > 0 $, so solutions in this interval increase. If $-1 < u < 1$ then $ f(u) < 0 $, so solutions in this interval decrease. If $ u > 1 $ then $ f(u) > 1 $ and solutions in this interval increase. This is summarized in the phase portrait of Figure 2.7; the arrows between the fixed points indicate whether solutions increase or decrease in that interval.### 2.3 Qualitative and Graphical Insights![Figure 2.7](#): Phase portrait for the ODE $ u' = u^2 - 1 $.Figure 2.7 shows a horizontal line (the $ u $-axis) marked with two blue dots at the fixed points $ u = -1 $ and $ u = 1 $. There are arrows between these fixed points indicating the direction of solution movement: arrows pointing right for $ u < -1 $ and $ u > 1 $, and arrows pointing left for $-1 < u < 1$.**Reading Exercise 2.3.4** Sketch a phase portrait for the autonomous ODE $ u'(t) = u^2(t) - 2u(t) - 3 $.

Answered: Image /qna-images/answer/12a1a93c-bf9e-48e4-b2c8-442a928e33e4.jpg

Exercise 2.3.3 For each ODE sketch a phase portrait by hand, following the procedure of Examples 2.10, 2.11, and 2.12. Classify each fixed point as asymptotically stable or unstable. Use the result to sketch solutions for the given initial conditions on pair of tu axes with a reasonable range for u. (a) u'(t)=-u(t), sketch solutions with u(0) = 2 and u(0) = -2. (b) v(t)=11-2v(t), sketch solutions with v(0) = 0 and v(0) = 15. H (c) v(t)=11-kv(t) (k a positive constant), sketch solutions with v(0) = 0 and v(0) = 15/k. (d) u' (t) = -(u(t)-1) (u(t)-3), sketch solutions with u(0) = 1/2, u(0) = 2, and u(0) = 4. (e) u' (t) = u(t)(1-u(t))-u(t)/10 (the harvested logistic equation (1.12) with r=1 = 1, K=1, and h=1/10), sketch solutions with u(0) = 1/2 and u(0) = 3/2. Note only u 20 makes physical sense here. What is the long-term fate of the species?

Exercise 2.3.3 For each ODE sketch a phase portrait by hand, following the procedure of Examples 2.10, 2.11, and 2.12. Classify each fixed point as asymptotically stable or unstable. Use the result to sketch solutions for the given initial conditions on pair of tu axes with a reasonable range for u. (a) u'(t)=-u(t), sketch solutions with u(0) = 2 and u(0) = -2. (b) v(t)=11-2v(t), sketch solutions with v(0) = 0 and v(0) = 15. H (c) v(t)=11-kv(t) (k a positive constant), sketch solutions with v(0) = 0 and v(0) = 15/k. (d) u' (t) = -(u(t)-1) (u(t)-3), sketch solutions with u(0) = 1/2, u(0) = 2, and u(0) = 4. (e) u' (t) = u(t)(1-u(t))-u(t)/10 (the harvested logistic equation (1.12) with r=1 = 1, K=1, and h=1/10), sketch solutions with u(0) = 1/2 and u(0) = 3/2. Note only u 20 makes physical sense here. What is the long-term fate of the species?

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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do b and c

**Exercise 2.3.3**

For each Ordinary Differential Equation (ODE), sketch a phase portrait by hand, following the procedure of Examples 2.10, 2.11, and 2.12. Classify each fixed point as asymptotically stable or unstable. Use the result to sketch solutions for the given initial conditions on a pair of $u$ axes with a reasonable range for $u$.

(a) $ u'(v) = -u(v) $, sketch solutions with $ u(0) = 2 $ and $ u(0) = -2 $.

(b) $ v'(u) = 11 - 2v(t) $, sketch solutions with $ v(0) = 0 $ and $ v(0) = 15 $.

(c) $ u'(v) = -ku(v) $ (a positive constant), sketch solutions with $ v(0) = 0 $ and $ v(0) = 15/k $.

(d) $ u'(v) = -(u(t) - 1)(u(t) - 3) $, sketch solutions with $ u(0) = 1/2 $, $ u(0) = 2 $, and $ u(0) = 4 $.

(e) $ u'(t) = r(u(t)(1 - u(t)) - u(t)/10 $ (the harvested logistic equation (1.12) with $ r = 1 $, $ K = 1 $, and $ h = 1/10 $), sketch solutions with $ u(0) = 1/2 $ and $ u(0) = 3/2 $. Note only $ u \geq 0 $ makes physical sense here. What is the long-term fate of the species?

(f) $ u'(t) = u(t)(1 - \frac{u(t)}{2}) - 2u(t) $ (the harvested logistic equation (1.12) with $ r = 1 $, $ K = 1 $, and $ h = 2 $), sketch solutions with $ u(0) = 1/2 $ and $ u(0) = 3/2 $. Note only $ u \geq 0 $ makes physical sense here. What is the long-term fate of

**Example 2.10** Let us sketch a phase portrait for the autonomous ODE $ u'(t) = u^2(t) - 1 $. In this case, the ODE is $ u' = f(u) $ with $ f(u) = u^2 - 1 $ and the fixed points are the solutions to $ f(u) = 0 $, that is, $ u^2 - 1 = 0 $. These fixed points are $ u = -1 $ and $ u = 1 $, shown as the blue dots on the $ u $-axis in Figure 2.7.

The fixed points divide the $ u $-axis into intervals $(-\infty, -1)$, $(-1, 1)$, and $(1, \infty)$. If $ u < -1 $ then $ f(u) > 0 $, so solutions in this interval increase. If $-1 < u < 1$ then $ f(u) < 0 $, so solutions in this interval decrease. If $ u > 1 $ then $ f(u) > 1 $ and solutions in this interval increase. This is summarized in the phase portrait of Figure 2.7; the arrows between the fixed points indicate whether solutions increase or decrease in that interval.

### 2.3 Qualitative and Graphical Insights

![Figure 2.7](#): Phase portrait for the ODE $ u' = u^2 - 1 $.

Figure 2.7 shows a horizontal line (the $ u $-axis) marked with two blue dots at the fixed points $ u = -1 $ and $ u = 1 $. There are arrows between these fixed points indicating the direction of solution movement: arrows pointing right for $ u < -1 $ and $ u > 1 $, and arrows pointing left for $-1 < u < 1$.

**Reading Exercise 2.3.4** Sketch a phase portrait for the autonomous ODE $ u'(t) = u^2(t) - 2u(t) - 3 $.

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