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2.2E: Separable Equations (Exercises)

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Q2.2.1

In Exercises 2.2.1-2.2.6 find all solutions.

1. y=3x2+2x+1y2

2. (sinx)(siny)+(cosy)y=0

3. xy+y2+y=0

4. yln|y|+x2y=0

5. (3y3+3ycosy+1)y+(2x+1)y1+x2=0

6. x2yy=(y21)3/2

Q2.2.2

In Exercises 2.2.7-2.2.10 find all solutions. Also, plot a direction field and some integral curves on the indicated rectangular region.

7. y=x2(1+y2);{1x1, 1y1}

8. y(1+x2)+xy=0;{2x2, 1y1}

9. y=(x1)(y1)(y2);{2x2, 3y3}

10. (y1)2y=2x+3;{2x2, 2y5}

Q2.2.3

In Exercises 2.2.11 and 2.2.12 solve the initial value problem.

11. y=x2+3x+2y2,y(1)=4

12. y+x(y2+y)=0,y(2)=1

Q2.2.4

In Exercises 2.2.13-2.2.16 solve the initial value problem and graph the solution.

13. (3y2+4y)y+2x+cosx=0,y(0)=1

14. y+(y+1)(y1)(y2)x+1=0,y(1)=0

15. y+2x(y+1)=0,y(0)=2

16. y=2xy(1+y2),y(0)=1

Q2.2.5

In Exercises 2.2.17-2.2.23 solve the initial value problem and find the interval of validity of the solution.

17. y(x2+2)+4x(y2+2y+1)=0,y(1)=1

18. y=2x(y23y+2),y(0)=3

19. y=2x1+2y,y(2)=0 &

20. y=2yy2,y(0)=1

21. x+yy=0,y(3)=4

22. y+x2(y+1)(y2)2=0,y(4)=2

23. (x+1)(x2)y+y=0,y(1)=3

Q2.2.6

24. Solve y=(1+y2)(1+x2) explicitly.

25. Solve y1x2+1y2=0 explicitly.

26. Solve y=cosxsiny,y(π)=π2 explicitly.

27. Solve the initial value problem y=ayby2,y(0)=y0. Discuss the behavior of the solution if a y00; b y0<0.

28. The population P=P(t) of a species satisfies the logistic equation P=aP(1αP) and P(0)=P0>0. Find P for t>0, and find limtP(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(SI), where r is a positive constant. Assuming that I(0)=I0, find I(t) for t>0, and show that limtI(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(SI)qI where q is a positive constant.

  1. Choose r and S positive. By plotting direction fields and solutions of (A) on suitable rectangular grids R={0tT, 0Id} in the (t,I)-plane, verify that if I is any solution of (A) such that I(0)>0, then limtI(t)=Sq/r if q<rS and limtI(t)=0 if qrS.
  2. To verify the experimental results of (a), use separation of variables to solve (A) with initial condition I(0)=I0>0, and find limtI(t).

31. Consider the differential equation y=ayby2q, where a, b are positive constants, and q is an arbitrary constant. Suppose y denotes a solution of this equation that satisfies the initial condition y(0)=y0.

  1. Choose a and b positive and q<a2/4b. By plotting direction fields and solutions of (A) on suitable rectangular grids R={0tT, cyd} in the (t,y)-plane, discover that there are numbers y1 and y2 with y1<y2 such that if y0>y1 then limty(t)=y2, and if y0<y1 then y(t)= for some finite value of t. (What happens if y0=y1?)
  2. Choose a and b positive and q=a2/4b. By plotting direction fields and solutions of (A) on suitable rectangular grids of the form (B), discover that there’s a number y1 such that if y0y1 then limty(t)=y1, while if y0<y1 then y(t)= for some finite value of t.
  3. Choose positive a, b and q>a2/4b. By plotting direction fields and solutions of (A) on suitable rectangular grids of the form (B), discover that no matter what y0 is, y(t)= for some finite value of t.
  4. Verify your results experiments analytically. Start by separating variables in (A) to obtain yayby2q=1. To decide what to do next you’ll have to use the quadratic formula. This should lead you to see why there are three cases. Take it from there! Because of its role in the transition between these three cases, q0=a2/4b is called a bifurcation value of q. In general, if q is a parameter in any differential equation, q0 is said to be a bifurcation value of q if the nature of the solutions of the equation with q<q0 is qualitatively different from the nature of the solutions with q>q0.

32. By plotting direction fields and solutions of y=qyy3, convince yourself that q0=0 is a bifurcation value of q for this equation. Explain what makes you draw this conclusion.

33. Suppose a disease spreads according to the model of Exercise 2.2.29, but there’s a medication that cures the infected population at a constant rate of q individuals per unit time, where q>0. Then the equation for the number of infected individuals becomes I=rI(SI)q.

Assuming that I(0)=I0>0, use the results of Exercise 2.2.31 to describe what happens as t.

34. Assuming that p, state conditions under which the linear equation y'+p(x)y=f(x) \nonumber is separable. If the equation satisfies these conditions, solve it by separation of variables and by the method developed in Section 2.1.

Q2.2.7

Solve the equations in Exercises 2.2.35-2.2.38 using variation of parameters followed by separation of variables.

35. {y'+y={2xe^{-x}\over1+ye^x}}&

36. {xy'-2y={x^6\over y+x^2}}

37. {y'-y}={(x+1)e^{4x}\over(y+e^x)^2}&

38. y'-2y= {xe^{2x}\over1-ye^{-2x}}

39. Use variation of parameters to show that the solutions of the following equations are of the form y=uy_1, where u satisfies a separable equation u'=g(x)p(u). Find y_1 and g for each equation.

  1. xy'+y=h(x)p(xy)
  2. {xy'-y=h(x) p\left({y\over x}\right)}
  3. y'+y=h(x) p(e^xy)
  4. xy'+ry=h(x) p(x^ry)
  5. {y'+{v'(x)\over v(x)}y= h(x) p\left(v(x)y\right)}

This page titled 2.2E: Separable Equations (Exercises) is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.

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