2.1E: Separable Equations (Exercises)
- Page ID
- 43273
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In Exercises 2.1.1-2.1.6 find all solutions.
1. \( {y'={3x^2+2x+1\over y-2}}\)&
2. \((\sin x)(\sin y)+(\cos y)y'=0\)
3. \(xy'+y^2+y=0\) &
4. \(y' \ln |y|+x^2y= 0\)
5. \( {(3y^3+3y \cos y+1)y'+{(2x+1)y\over 1+x^2}=0}\)
6. \(x^2yy'=(y^2-1)^{3/2}\)
Q2.1.2
In Exercises 2.1.7-2.1.10 find all solutions. Also, plot a direction field and some integral curves on the indicated rectangular region.
7. \( {y'=x^2(1+y^2)}; \; \{-1\le x\le1,\ -1\le y\le1\}\)
8. \(y'(1+x^2)+xy=0 ; \; \{-2\le x\le2,\ -1\le y\le1\}\)
9. \(y'=(x-1)(y-1)(y-2); \; \{-2\le x\le2,\ -3\le y\le3\}\)
10. \((y-1)^2y'=2x+3; \; \{-2\le x\le2,\ -2\le y\le5\}\)
Q2.1.3
In Exercises 2.1.11 and 2.1.12 solve the initial value problem.
11. \( {y'={x^2+3x+2\over y-2}, \quad y(1)=4}\)
12. \(y'+x(y^2+y)=0, \quad y(2)=1\)
Q2.1.4
In Exercises 2.1.13-2.1.16 solve the initial value problem and graph the solution.
13. \((3y^2+4y)y'+2x+\cos x=0, \quad y(0)=1\)
14. \( {y'+{(y+1)(y-1)(y-2)\over x+1}=0, \quad y(1)=0}\)
15. \(y'+2x(y+1)=0, \quad y(0)=2\)
16. \(y'=2xy(1+y^2),\quad y(0)=1\)
Q2.1.5
In Exercises 2.1.17-2.1.23 solve the initial value problem and find the interval of validity of the solution.
17. \(y'(x^2+2)+ 4x(y^2+2y+1)=0, \quad y(1)=-1\)
18. \(y'=-2x(y^2-3y+2), \quad y(0)=3\)
19. \( {y'={2x\over 1+2y}, \quad y(2)=0}\) &
20. \(y'=2y-y^2, \quad y(0)=1\)
21. \(x+yy'=0, \quad y(3) =-4\)
22. \(y'+x^2(y+1)(y-2)^2=0, \quad y(4)=2\)
23. \((x+1)(x-2)y'+y=0, \quad y(1)=-3\)
Q2.1.6
24. Solve \( {y'={(1+y^2) \over (1+x^2)}}\) explicitly.
25. Solve \( {y'\sqrt{1-x^2}+\sqrt{1-y^2}=0}\) explicitly.
26. Solve \( {y'={\cos x\over \sin y},\quad y (\pi)={\pi\over2}}\) explicitly.
27. Solve the initial value problem \[y'=ay-by^2,\quad y(0)=y_0.\] Discuss the behavior of the solution if a \(y_0\ge0\); b \(y_0<0\).
28. The population \(P=P(t)\) of a species satisfies the logistic equation \[P'=aP(1-\alpha P)\] and \(P(0)=P_0>0\). Find \(P\) for \(t>0\), and find \(\lim_{t\to\infty}P(t)\).
29. An epidemic spreads through a population at a rate proportional to the product of the number of people already infected and the number of people susceptible, but not yet infected. Therefore, if \(S\) denotes the total population of susceptible people and \(I=I(t)\) denotes the number of infected people at time \(t\), then \[I'=rI(S-I),\] where \(r\) is a positive constant. Assuming that \(I(0)=I_0\), find \(I(t)\) for \(t>0\), and show that \(\lim_{t\to\infty}I(t)=S\).
30. The result of Exercise 2.2.29 is discouraging: if any susceptible member of the group is initially infected, then in the long run all susceptible members are infected! On a more hopeful note, suppose the disease spreads according to the model of Exercise 2.2.29, but there’s a medication that cures the infected population at a rate proportional to the number of infected individuals. Now the equation for the number of infected individuals becomes \[I'=rI(S-I)-qI \tag{A} \] where \(q\) is a positive constant.
- Choose \(r\) and \(S\) positive. By plotting direction fields and solutions of (A) on suitable rectangular grids \[R=\{0\le t \le T,\ 0\le I \le d\}\] in the \((t,I)\)-plane, verify that if \(I\) is any solution of (A) such that \(I(0)>0\), then \(\lim_{t\to\infty}I(t)=S-q/r\) if \(q<rS\) and \(\lim_{t\to\infty}I(t)=0\) if \(q\ge rS\).
- To verify the experimental results of (a), use separation of variables to solve (A) with initial condition \(I(0)=I_0>0\), and find \(\lim_{t\to\infty}I(t)\).