Direction Fields and Numerical Methods

연습 (40)

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1. Create a direction field for the differential equation \(y'={x}^{2}-{y}^{2}\) and sketch a solution curve passing through the point \((-1,2).\)

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2. Create a direction field for the differential equation \(y'={(y-3)}^{2}({y}^{2}+y-2)\) and identify any equilibrium solutions. Classify each of the equilibrium solutions as stable, unstable, or semi-stable.

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   The direction field is

   The equilibrium solutions are \(y=-2,y=1,\) and \(y=3.\) To classify each of the solutions, look at an arrow directly above or below each of these values. For example, at \(y=-2\) the arrows directly below this solution point up, and the arrows directly above the solution point down. Therefore all initial conditions close to \(y=-2\) approach \(y=-2,\) and the solution is stable. For the solution \(y=1,\) all initial conditions above and below \(y=1\) are repelled (pushed away) from \(y=1,\) so this solution is unstable. The solution \(y=3\) is semi-stable, because for initial conditions slightly greater than \(3,\) the solution approaches infinity, and for initial conditions slightly less than \(3,\) the solution approaches \(y=3.\)

3. Create a direction field for the differential equation \(y'=(x+5)(y+2)({y}^{2}-4y+4)\) and identify any equilibrium solutions. Classify each of the equilibrium solutions as stable, unstable, or semi-stable.

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   The equilibrium solutions are \(y=-2\) and \(y=2.\) For this equation, \(y=-2\) is an unstable equilibrium solution, and \(y=2\) is a semi-stable equilibrium solution.

4. Consider the initial-value problem

   \[{y}^{'}=3{x}^{2}-{y}^{2}+1,\ y(0)=2.\]

   Use Euler’s method with a step size of \(0.1\) to generate a table of values for the solution for values of \(x\) between \(0\) and \(1.\)

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   We are given \(h=0.1\) and \(f(x,y)=3{x}^{2}-{y}^{2}+1.\) Furthermore, the initial condition \(y(0)=2\) gives \({x}_{0}=0\) and \({y}_{0}=2.\) Using with \(n=0,\) we can generate .

   \(n\)	\({x}_{n}\)	\({y}_{n}={y}_{n-1}+hf({x}_{n-1},{y}_{n-1})\)
   \(0\)	\(0\)	\(2\)
   \(1\)	\(0.1\)	\({y}_{1}={y}_{0}+hf({x}_{0},{y}_{0})=1.7\)
   \(2\)	\(0.2\)	\({y}_{2}={y}_{1}+hf({x}_{1},{y}_{1})=1.514\)
   \(3\)	\(0.3\)	\({y}_{3}={y}_{2}+hf({x}_{2},{y}_{2})=1.3968\)
   \(4\)	\(0.4\)	\({y}_{4}={y}_{3}+hf({x}_{3},{y}_{3})=1.3287\)
   \(5\)	\(0.5\)	\({y}_{5}={y}_{4}+hf({x}_{4},{y}_{4})=1.3001\)
   \(6\)	\(0.6\)	\({y}_{6}={y}_{5}+hf({x}_{5},{y}_{5})=1.3061\)
   \(7\)	\(0.7\)	\({y}_{7}={y}_{6}+hf({x}_{6},{y}_{6})=1.3435\)
   \(8\)	\(0.8\)	\({y}_{8}={y}_{7}+hf({x}_{7},{y}_{7})=1.4100\)
   \(9\)	\(0.9\)	\({y}_{9}={y}_{8}+hf({x}_{8},{y}_{8})=1.5032\)
   \(10\)	\(1.0\)	\({y}_{10}={y}_{9}+hf({x}_{9},{y}_{9})=1.6202\)

   With ten calculations, we are able to approximate the values of the solution to the initial-value problem for values of \(x\) between \(0\) and \(1.\)

5. Consider the initial-value problem

   \[{y}^{'}={x}^{3}+{y}^{2},\ y(1)=-2.\]

   Using a step size of \(0.1,\) generate a table with approximate values for the solution to the initial-value problem for values of \(x\) between \(1\) and \(2.\)

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   \(n\)	\({x}_{n}\)	\({y}_{n}={y}_{n-1}+hf({x}_{n-1},{y}_{n-1})\)
   \(0\)	\(1\)	\(-2\)
   \(1\)	\(1.1\)	\({y}_{1}={y}_{0}+hf({x}_{0},{y}_{0})=-1.5\)
   \(2\)	\(1.2\)	\({y}_{2}={y}_{1}+hf({x}_{1},{y}_{1})=-1.1419\)
   \(3\)	\(1.3\)	\({y}_{3}={y}_{2}+hf({x}_{2},{y}_{2})=-0.8387\)
   \(4\)	\(1.4\)	\({y}_{4}={y}_{3}+hf({x}_{3},{y}_{3})=-0.5487\)
   \(5\)	\(1.5\)	\({y}_{5}={y}_{4}+hf({x}_{4},{y}_{4})=-0.2442\)
   \(6\)	\(1.6\)	\({y}_{6}={y}_{5}+hf({x}_{5},{y}_{5})=0.0993\)
   \(7\)	\(1.7\)	\({y}_{7}={y}_{6}+hf({x}_{6},{y}_{6})=0.5099\)
   \(8\)	\(1.8\)	\({y}_{8}={y}_{7}+hf({x}_{7},{y}_{7})=1.0272\)
   \(9\)	\(1.9\)	\({y}_{9}={y}_{8}+hf({x}_{8},{y}_{8})=1.7159\)
   \(10\)	\(2\)	\({y}_{10}={y}_{9}+hf({x}_{9},{y}_{9})=2.6962\)

6. Are there any equilibria? What are their stabilities?

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   \(y=0\) is a stable equilibrium

7. Are there any equilibria? What are their stabilities?

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   \(y=0\) is a stable equilibrium and \(y=2\) is unstable

8. \(y'={t}^{3}\)

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9. \(y'={e}^{t}\)

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   General solution is \(y={e}^{t}+C\).

10. \(\frac{dy}{dx}={x}^{2}\text{cos}\ x\)

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11. \(\frac{dy}{dt}=t{e}^{t}\)

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    General solution is \(y={e}^{t}(t-1)+C\).

12. \(\frac{dx}{dt}=\text{cosh}(t)\)

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13. \(y'={y}^{2}-1\)

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14. \(y'=1-{y}^{2}-{x}^{2}\)

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15. \(y'={t}^{2}\text{sin}\ y\)

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16. \(y'=3y+xy\)

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17. \(y'={e}^{t}\)

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18. \(y'=\frac{1}{2}y+t\)

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    A

19. \(y'=\text{-}ty\)

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20. \(y'=t\ \text{sin}\ y\)

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    B

21. \(y'=\text{-}t\ \text{cos}\ y\)

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22. \(y'=t\ \text{tan}\ y\)

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    A

23. \(y'={\text{sin}}^{2}y\)

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24. \(y'={y}^{2}{t}^{3}\)

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    C

25. \(y'=-3y,\ y(0)=1\)

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26. \(y'={t}^{2},y(0)=2\)

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    \(2.24,\) exact: \(2\)

27. \({y}^{'}=3t-y,y(0)=1.\) Exact solution is \(y=3t+4{e}^{\text{-}t}-3\)

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28. \({y}^{'}=y+{t}^{2},y(0)=3.\) Exact solution is \(y=5{e}^{t}-2-{t}^{2}-2t\)

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    \(7.739264,\) exact: \(5(e-1)\)

29. \({y}^{'}=2t,y(0)=0\)

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30. [T] \(y'={e}^{(x+y)},y(0)=-1.\) Exact solution is \(y=\text{-}\text{ln}(e+1-{e}^{x})\)

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    \(-0.2535\) exact: \(0\)

31. \({y}^{'}={y}^{2}\text{ln}(x+1),y(0)=1.\) Exact solution is \(y=-\frac{1}{(x+1)(\text{ln}(x+1)-1)}\)

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32. \({y}^{'}={2}^{x},y(0)=0,\) Exact solution is \(y=\frac{{2}^{x}-1}{\text{ln}(2)}\)

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    \(1.345,\) exact: \(\frac{1}{\text{ln}(2)}\)

33. \({y}^{'}=y,y(0)=-1.\) Exact solution is \(y=\text{-}{e}^{x}.\)

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34. \({y}^{'}=-5t,y(0)=-2.\) Exact solution is \(y=-\frac{5}{2}{t}^{2}-2\)

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    \(-4,\) exact: \(\text{-}1\text{/}2\)

35. Show that, by our assumption that the total population size is constant \((S+I=N),\) you can reduce the system to a single differential equation in \(I\text{:}\ I'=c(N-I)I-rI.\)

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36. Assuming the parameters are \(c=0.5,N=5,\) and \(r=0.5,\) draw the resulting directional field.

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37. [T] Use computational software or a calculator to compute an approximate solution to the initial-value problem \(y'=ty,\ y(0)=2\) using Euler's Method with a given step size \(h.\) Find the solution at \(t=1\) in terms of \(h.\) Hint: Here is "pseudo-code" for how to write a computer program to perform Euler's Method for \(y'=f(t,y),y(0)={y}_{0}\text{:}\)

    • Create function \(f(t,y).\)

    • Define parameters \(y(0)={y}_{0},\ t(0)=0,\) step size \(h,\) and total number of steps, \(N\)

    • Write a for loop:

      for \(\text{k}=0\ \text{to}\ N\)

      \(\ \text{fn}=\text{f}(\text{t}(\text{k}),\text{y}(\text{k}))\)

      \(\ \text{y}(\text{k}+\text{1})=\text{y}(\text{k})+\text{h*fn}\)

      \(\ \text{t}(\text{k}+\text{1})=\text{t}(\text{k})+\text{h}\)

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38. Solve the initial-value problem for the exact solution.

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    \(y'=2{e}^{{t}^{2}\text{/}2}\)

39. Draw the directional field

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40. Compute using \(h=1\)

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    \(2\)

이 페이지의 일부는 다음에서 변경되었습니다. OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). 여기서 압축하고 다시 설명; 오류는 우리의.

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