1. 1
2. 3
3. 1
4. 2
5. 1

3. $y=-\log(c-x^2).$

5. $\displaystyle \frac{e^y-1}{e^y}=k\exp\left(\frac{x^2}{2}\right).$

7. $\displaystyle \frac{y-3}{y+3}=-\frac12e^{6x}.$

9. $\tan(y)=x.$

11. $y=kx^4$.

13. $e^2$

15. $2/3$

17. $-4/3$

19. $\displaystyle\frac{dy}{dx}=2xy$, $y(0)=1.$

21. $\displaystyle\frac{dy}{dx}=-\exp(x-y).$

23.

1. $y_1(h)=1+h$
2. $$\lim_{h\to 0}\frac{-h-1+2\exp(h)-(1+h)}{h^2}=\lim_{h\to 0}\frac{-2-2h+2\exp(h)}{h^2} = 1$$ after two applications of L'Hôpital's rule (you could also write out $\exp(h)$ as a series and check the lower-order terms cancel).

25.

1. $y_1(h)=1+h+h^2$
2. $$\lim_{h\to 0}\frac{-h-1+2\exp(h)-(1+h+h^2h)}{h^3}=\lim_{h\to 0}\frac{-2-2h-h^2+2\exp(h)}{h^3} = \frac{8}{3}$$ after three applications of L'Hôpital's rule (you could also write out $\exp(h)$ as a series and check the lower-order terms cancel).

27.

1. Both
2. Linear
3. Linear
4. Exact

29. $\displaystyle y=\exp(x)-\frac{\exp(x)}{x}+\frac{C}{x}.$

31. $y=1-x+C\exp(-x).$

33. $y=(1/3)\exp(x)+(5/3)\exp(-2x).$

35. $2x^3+6x^2y-3y^2=-3.$

37. After 120 minutes, the concentration tank 1 is $2(1-\exp(-0.6)) \approx 0.90 mg/l$. The concentration in tank 2 is $2-(2+t/100)\exp(-0.6) \approx 0.24 mg/l$.

39. 0.00092 g/l or 0.9mg/l, so tank 2 has a max concentration less than 2% of the max concentration in tank 1. One of the advantages of putting tanks in series is "resistance to shocks" since contamination in later tanks will be much lower than in the initial tank.

41. $y(2)=1.$

43. $y(1)\approx 3.5194.$

45. $y(2)\approx -3.2473.$

47. $x=C\exp(-2y)+\exp(-y).$

49. Multiplying through by $(x+y)^r$ gives us $(x+4y)(x+y)^r dx + (3x+6y)(x+y)^r dy = 0$. Noting that $x+4y=(x+y)+3y$ and $3x+6y=3(x+y)+3y$, computing the partial derivatives in the test for exactness gives: $$\begin{align} \frac{\partial}{\partial y}\left((x+4y)(x+y)^r\right) &= \frac{\partial}{\partial y} \left((x+y)^{r+1}+3y(x+y)^r\right) \\ &=(r+1)(x+y)^r + 3(x+y)^3 + 3ry(x+y)^{r-1} &= (r+4)(x+y)^r + 3ry(x+y)^{r-1}. \\ \frac{\partial}{\partial x}\left((3x+6y)(x+y)^r\right) \\ &= \frac{\partial}{\partial x}\left(3(x+y)^{r+1} + 3y(x+y)^r\right) \\ &=3(r+1)(x+y)^r + 3ry(x+y)^{r-1}. \end{align}$$ Comparing the two partial derivatives we see they are equal when $3(r+1)=r+4$, which gives us $r=1/2$. So $(x+y)^{1/2}$ is an integrating factor.

51. $$\begin{align} \frac{dw}{dt}+aw & = \frac{d(x-\tilde{x})}{dt}+a(x-a\tilde{x}) \\ &=\left(\frac{dx}{dt}+ax\right) - \left(\frac{d\tilde{x}}{dt}+a\tilde{x}\right) \\ &=f(t) - \tilde{f}(t)\end{align}$$ and $w(0)=x(0)-\tilde{x}(0)=x_0-x_0=0$.

53.

1. Homogeneous
2. Exact
3. None of the above
4. Linear

55. $\displaystyle y= \frac{1}{C\exp(2x)+\exp(x)}$ and singular solution $y=0.$

57. $y=2x.$

59. $\displaystyle y=\frac{x}{C\exp(2x)-x/2-1/4}+1.$

61. $\displaystyle y=\exp(x)-\frac{1}{C\exp(2x)+\frac13\exp(-x)}.$

63. $\displaystyle y=\frac{x^2-x-1}{x-1}.$

65.

1. $y=0$ is an unstable equilibrium.
   $y=1$ is a stable equilibrium.
   $y=3$ is an unstable equilibrium.

2. $y=-3$ is a stable equilibrium.
   $y=-1$ is a semi-stable equilibrium.
   $y=1$ is an unstable equilibrium.

67. Along the line $y=x,$ the slope is $\displaystyle \frac{dy}{dx}=x^2-y^2 = x^2 - x^2 = 0.$ Along the line $y=0$ the slope is $\displaystyle \frac{dy}{dx}=x^2-y^2 = x^2-0=x^2>0.$ Drawing the slope field we see that the arrows all point into the wedge formed by those two lines in the first quadrant. So since we start at the edge of the wedge at (1,0), we stay inside the wedge and the solution therefore satisfies $0 < y(x) < x.$

69.

1. $\displaystyle p=12500\left(1-\sqrt{1-\frac{H}{1250}}\right)$ is unstable.
   $\displaystyle p=12500\left(1+\sqrt{1-\frac{H}{1250}}\right)$ is stable.

2. The maximum $H$ which has a semi-stable equilibrium is 1250. Any smaller $H$ has a stable equilibrium.

71.

1. The equilibria are $p=0$ and $\displaystyle p=\left(1-\frac{E}{r}\right)C.$

2. $p=0$ is unstable and $\displaystyle p=\left(1-\frac{E}{r}\right)C$ is stable.

73. $y=cx+1/c$ for $c\ne0$. The singular solution (which was not required) is $y=\pm2\sqrt{x}$ or $y^2=4x$ if you prefer.

75. $y=(x+1)y'-(y')^2$.

77. This is not a solution to the differential equation because it is not differentiable at $x=0$.

79. Since the envelope is tangent to a solution curve at every point, that means $x$, $y$, and $y'$ for the enveloping curve match those vales for a solution curve. Since any solution curve satisfies $F(x,y,y')=0$, so does the envelope, hence the envelope also solves the differential equation.

81.

1. $y_1(x)=x+1$.

2. $y_2(x)=x^2/2 + x + 1$