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. $e^2$

13. $2/3$

15. $-4/3$

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

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

21.

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

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

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

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

29. $2x^3+6x^2y-3y^2=-3.$

31. $y(2)=1.$

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

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

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

39. Moving everthing to the same side of the equation we have $M(x)dx-N(y)dy=0.$ The test for exactness then gives $\displaystyle \frac{\partial M}{\partial y} = 0 = \frac{\partial N}{\partial x}$ since $M(x)$ is a function of $x$ alone and $N(y)$ is a function of $y$ alone. So any separable equation is exact (once it is separated).

41.

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

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

45. $y=2x.$

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

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

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

53.

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.

55. 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 $0 < y(x) < x.$

57.

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.

59.

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.

61. Wolfram Alpha finds the general solution to be $\displaystyle y(x) = \frac{2\exp(c_1+x)-3}{\exp(c_1+x)-1}.$ It apparently neglects that there is an absolute value sign in the integral, $\displaystyle \int \frac{dy}{y} = \log|y|+C$, so it doesn't adjust to have $\exp(c_1+x)=k\exp(x)$ where $k=\pm\exp(c_1)$, as we discussed in the section on separable equations. Therefore, when it gets an initial value that corresponds to a negative or zero value of $k$, it can't find the answer. It also misses the singular solution of $y=2$. There is still a place for a thinking person instead of a thinking machine in working with these sorts of problems :)