2.4E: Transformation of Nonlinear Equations into Separable Equations (Exercises)

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- 2.4: Transformation of Nonlinear Equations into Separable Equations
- 2.5: Exact Equations

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Q2.4.1

In Exercises 2.4.1-2.4.4 solve the given Bernoulli equation.

1. $y'+y=y^2$

2. ${7xy'-2y=-{x^2 \over y^6}}$

3. $x^2y'+2y=2e^{1/x}y^{1/2}$

4. ${(1+x^2)y'+2xy ={1 \over (1+x^2)y}}$

Q2.4.2

In Exercises 2.4.5 and 2.4.6 find all solutions. Also, plot a direction field and some integral curves on the indicated rectangular region.

5. $y'-xy=x^3y^3; \quad \{-3\le x\le 3, -2\le y\le 2\}$

6. ${y'-{1+x\over 3x}y=y^4}; \quad \{-2\le x\le2,-2\le y \le2\}$

Q2.4.3

In Exercises 2.4.7-2.4.11 solve the initial value problem.

7. $y'-2y=xy^3,\quad y(0)=2\sqrt2$

8. $y'-xy=xy^{3/2},\quad y(1)=4$

9. $xy'+y=x^4y^4,\quad y(1)=1/2$

10. $y'-2y=2y^{1/2},\quad y(0)=1$

11. ${y'-4y={48x\over y^2},\quad y(0)=1}$

Q2.4.4

In Exercises 2.4.12 and 2.4.13 solve the initial value problem and graph the solution.

12. $x^2y'+2xy=y^3,\quad y(1)=1/\sqrt2$

13. $y'-y=xy^{1/2},\quad y(0)=4$

Q2.4.5

14. You may have noticed that the logistic equation $P'=aP(1-\alpha P) \nonumber$ from Verhulst’s model for population growth can be written in Bernoulli form as $P'-aP=-a\alpha P^2. \nonumber$ This isn’t particularly interesting, since the logistic equation is separable, and therefore solvable by the method studied in Section 2.2. So let’s consider a more complicated model, where $a$ is a positive constant and $\alpha$ is a positive continuous function of $t$ on $[0,\infty)$ . The equation for this model is $P'-aP=-a\alpha(t) P^2, \nonumber$ a non-separable Bernoulli equation.

Assuming that $P(0)=P_0>0$ , find $P$ for $t>0$ .
Verify that your result reduces to the known results for the Malthusian model where $\alpha=0$ , and the Verhulst model where $\alpha$ is a nonzero constant.
Assuming that $\lim_{t\to\infty}e^{-at}\int_0^t\alpha(\tau)e^{a\tau}\,d\tau=L \nonumber$ exists (finite or infinite), find $\lim_{t\to\infty}P(t)$ .

Q2.4.6

In Exercises 2.4.15-2.4.18 solve the equation explicitly.

15. $y'= {y+x\over x}$

16. $y'= {y^2+2xy \over x^2}$

17. $xy^3y'=y^4+x^4$

18. $y'= {y\over x}+\sec{y\over x}$

Q2.4.7

In Exercises 2.4.19-2.4.21 solve the equation explicitly. Also, plot a direction field and some integral curves on the indicated rectangular region.

19. $x^2y'=xy+x^2+y^2; \quad \{-8\le x\le 8,-8\le y\le 8\}$

20. $xyy'=x^2+2y^2; \quad \{-4\le x\le 4,-4\le y\le 4\}$

21. $y'= {2y^2+x^2e^{-(y/x)^2}\over 2xy}; \quad \{-8\le x\le 8,-8\le y\le 8\}$

Q2.4.8

In Exercises 2.4.22-2.4.27 solve the initial value problem.

22. $y'= {xy+y^2\over x^2}, \quad y(-1)=2$

23. $y'= {x^3+y^3\over xy^2}, \quad y(1)=3$

24. $xyy'+x^2+y^2=0, \quad y(1)=2$

25. $y'= {y^2-3xy-5x^2 \over x^2}, \quad y(1)=-1$

26. $x^2y'=2x^2+y^2+4xy, \quad y(1)=1$

27. $xyy'=3x^2+4y^2, \quad y(1)=\sqrt{3}$

Q2.4.9

In Exercises 2.4.28-2.4.34 solve the given homogeneous equation implicitly.

28. $y'= {x+y \over x-y}$

29. $(y'x-y)(\ln |y|-\ln |x|)=x$

30. $y'= {y^3+2xy^2+x^2y+x^3\over x(y+x)^2}$

31. $y'= {x+2y \over 2x+y}$

32. $y'= {y \over y-2x}$

33. $y'= {xy^2+2y^3\over x^3+x^2y+xy^2}$

$y'= {x^3+x^2y+3y^3 \over x^3+3xy^2}$

Q2.4.10

35.

Find a solution of the initial value problem $x^2y'=y^2+xy-4x^2, \quad y(-1)=0 \tag{A}$ on the interval $(-\infty,0)$ . Verify that this solution is actually valid on $(-\infty,\infty)$ .
Use Theorem 2.3.1 to show that (A) has a unique solution on $(-\infty,0)$ .
Plot a direction field for the differential equation in (A) on a square $\{-r\le x\le r, -r\le y\le r\}, \nonumber$ where $r$ is any positive number. Graph the solution you obtained in (a) on this field.
Graph other solutions of (A) that are defined on $(-\infty,\infty)$ .
Graph other solutions of (A) that are defined only on intervals of the form $(-\infty,a)$ , where is a finite positive number.

36.

Solve the equation $xyy'=x^2-xy+y^2 \tag{A}$ implicitly.
Plot a direction field for (A) on a square $\{0\le x\le r,0\le y\le r\} \nonumber$ where $r$ is any positive number.
Let $K$ be a positive integer. (You may have to try several choices for $K$ .) Graph solutions of the initial value problems $xyy'=x^2-xy+y^2,\quad y(r/2)={kr\over K}, \nonumber$ for $k=1$ , $2$ , …, $K$ . Based on your observations, find conditions on the positive numbers $x_0$ and $y_0$ such that the initial value problem $xyy'=x^2-xy+y^2,\quad y(x_0)=y_0, \tag{B}$ has a unique solution (i) on $(0,\infty)$ or (ii) only on an interval $(a,\infty)$ , where $a>0$ ?
What can you say about the graph of the solution of (B) as $x\to\infty$ ? (Again, assume that $x_0>0$ and $y_0>0$ .)

37.

Solve the equation $y'={2y^2-xy+2x^2 \over xy+2x^2} \tag{A}$ implicitly.
Plot a direction field for (A) on a square $\{-r\le x\le r,-r\le y\le r\} \nonumber$ where $r$ is any positive number. By graphing solutions of (A), determine necessary and sufficient conditions on $(x_0,y_0)$ such that (A) has a solution on (i) $(-\infty,0)$ or (ii) $(0,\infty)$ such that $y(x_0)=y_0$ .

38. Follow the instructions of Exercise 2.4.37 for the equation $y'={xy+x^2+y^2 \over xy}. \nonumber$

39. Pick any nonlinear homogeneous equation $y'=q(y/x)$ you like, and plot direction fields on the square $\{-r\le x\le r,\ -r\le y\le r\}$ , where $r>0$ . What happens to the direction field as you vary $r$ ? Why?

40. Prove: If $ad-bc\ne 0$ , the equation $y'={ax+by+\alpha \over cx+dy+\beta} \nonumber$ can be transformed into the homogeneous nonlinear equation ${dY \over dX}={aX+bY \over cX+dY} \nonumber$ by the substitution $x=X-X_0,\ y=Y-Y_0$ , where $X_0$ and $Y_0$ are suitably chosen constants.

Q2.4.11

In Exercises 2.4.21-2.4.43 use a method suggested by Exercise 2.4.40 to solve the given equation implicitly.

41. $y'= {-6x+y-3 \over 2x-y-1}$

42. $y'= {2x+y+1 \over x+2y-4}$

43. $y'= {-x+3y-14 \over x+y-2}$

Q2.4.12

In Exercises 2.4.44-2.4.51 find a function $y_{1}$ such that the substitution $y = uy_{1}$ transforms the given equation into a separable equation of the form (2.4.6). Then solve the given equation explicitly.

44. $3xy^2y'=y^3+x$

45. $xyy'=3x^6+6y^2$

46. $x^3y'=2(y^2+x^2y-x^4)$

47. $y'=y^2e^{-x}+4y+2e^x$

48. $y'= {y^2+y\tan x+\tan^2 x\over\sin^2x}$

49. $x(\ln x)^2y'=-4(\ln x)^2+y\ln x+y^2$

50. $2x(y+2\sqrt x)y'=(y+\sqrt x)^2$

51. $(y+e^{x^2})y'=2x(y^2+ye^{x^2}+e^{2x^{2}}$

Q2.4.13

52. Solve the initial value problem $y'+{2\over x}y={3x^2y^2+6xy+2\over x^2(2xy+3)},\quad y(2)=2. \nonumber$

53. Solve the initial value problem $y'+{3\over x}y={3x^4y^2+10x^2y+6\over x^3(2x^2y+5)},\quad y(1)=1. \nonumber$

54. Prove: If $y$ is a solution of a homogeneous nonlinear equation $y'=q(y/x)$ , so is $y_1=y(ax)/a$ , where $a$ is any nonzero constant.

55. A generalized Riccati equation is of the form $y'=P(x)+Q(x)y+R(x)y^2. \tag{A}$ (If $R\equiv-1$ , (A) is a Riccati equation.) Let $y_1$ be a known solution and $y$ an arbitrary solution of (A). Let $z=y-y_1$ . Show that $z$ is a solution of a Bernoulli equation with $n=2$ .

Q2.4.14

In Exercises 2.4.56-2.4.59, given that $y_{1}$ is a solution of the given equation, use the method suggested by Exercise 2.4.55 to find other solutions.

56. $y'=1+x - (1+2x)y+xy^2$ ; $y_1=1$

57. $y'=e^{2x}+(1-2e^x)y+y^2$ ; $y_1=e^x$

58. $xy'=2-x+(2x-2)y-xy^2$ ; $y_1=1$

59. $xy'=x^3+(1-2x^2)y+xy^2$ ; $y_1=x$

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