3.

1. $\displaystyle f_e(x)+f_o(x)=\frac{f(x)+f(-x)}2 + \frac{f(x)-f(-x)}2 = \frac{2f(x)}2=f(x)$

2. $\displaystyle f_e(-x)=\frac{f(-x)+f(x)}2=\frac{f(x)+f(-x)}2 = f_e(x)$
   $\displaystyle f_o(-x)=\frac{f(-x)-f(x)}2=-\frac{f(x)-f(-x)}2 = -f_o(x)$

5.

1. Remember that $\cos(-x)=\cos(x)$ and $\sin(-x)=-\sin(x)$. $$ \begin{align} \cosh(ix)&=\frac{\exp(ix)+\exp(-ix)}2 \\ &=\frac{(\cos(x)+i\sin(x))+(\cos(-x)+i\sin(-x)}2 \\ &=\frac{\cos(x)+i\sin(x)+\cos(x)-i\sin(x)}2 \\ &=\frac{2\cos(x)}2 = \cos(x) \end{align} $$
2. $$ \begin{align} \sinh(ix)&=\frac{\exp(ix)-\exp(-ix)}2 \\ &=\frac{(\cos(x)+i\sin(x))-(\cos(-x)+i\sin(-x)}2 \\ &=\frac{\cos(x)+i\sin(x)-\cos(x)+i\sin(x)}2 \\ &=\frac{2i\sin(x)}2 = i\sin(x) \end{align} $$

7. $\ln(x+\sqrt{1+x^2}).$

9. $\sinh^{-1}(x)+C.$ If you use trig substitutions, you end up integrating $\sec(u)$ to $\ln(\sec(u)+\tan(u))+C$ and after back-substituting for $u$ you get the formula for $\sinh^{-1}(x)$ found above in problem 7.

11. $12\exp(2x).$

13. $12x^5.$

15. Remember that the differntiation operator $D$ is linear by the basic rules of differentiation from calculus 1. We first check $L(y+z)=Ly+Lz$. $$ \begin{align} L(y+z)&=D^2(y+z)+3D(y+z)-4(y+z) \\ &=D^2y+D^2z+3Dy+3Dz-4y-4z \\ &=(D^2y+3Dy-4y) + (D^2z+3Dz-4z) \\ &=Ly+Lz \end{align} $$ Second we check $L(cy)=cLy$. $$ \begin{align} L(cy)&=D^2(cy)+3D(cy)-4(cy) \\ &=c(D^2y)+c(3Dy)-c(4y) \\ &=c(D^2y+3Dy-4y) \\ &=c(Ly) \end{align} $$

17. Almost any choice of $y$ and $z$ will show this is not linear. For example, let $y=x$ and $z=1$. Then we compute $$ \begin{align} Ly&=x(Dx)=x \\ Lz&=1(D1)=0 \\ L(y+z)&=(x+1)D(x+1)=x+1 \end{align} $$ So $Ly+Lz=x+0\ne x+1=L(y+z).$

19. Computing the Wronskian we get $W(\exp(x),\exp(2x))=2\exp(x)\exp(2x)-\exp(x)\exp(2x) = \exp(3x).$ Since $W(0)=1\ne0$, the Wronskian is not identically 0 and hence the functions are linearly independent.

21. Computing the Wronskian we get $W(\exp(rx),x\exp(rx))=\exp(rx)((1+rx)\exp(rx))-rx\exp(rx)\exp(rx) =\exp(2rx).$ Since $W(0)=1\ne0$, the Wronskian is not identically 0 and hence the functions are linearly independent.

23.

1. b
2. a
3. c

25.

1. $y''+(x-1)y'-xy=0$
2. $y''+(x-1)y'+(1-x)y=0$

27.

1. $y''+(a(x)+b(x))y'+(a(x)b(x)+b'(x))y=0$
2. $y''+(a(x)+b(x))y'+(a(x)b(x)+a'(x))y=0$

29. $$ y=C_1\exp(-x)\int_0^x \exp(-t^2+t) dt + C_2\exp(-x)$$ 31. $$y = \exp(-x^2)\int_0^x\exp(t^2+t)dt + C_1\exp(-x^2)\int_0^x\exp(t^2-t)dt + C_2\exp(-x^2) $$ 33. $y=C_1\exp(rx)+C_2x\exp(rx)$.

35. $y_p=A\exp(2x)+Bx^2+Cx+D$ where $A,$ $B,$ $C,$ and $D$ are constants.

37. $y_p=Ax\sin(2x)+Bx\cos(2x)+Cx^2\exp(x)+Dx\exp(x)+E\exp(x)$ where $A,$ $B,$ $C,$ $D,$ and $E$ are constants.

39. $y=C_1\exp((1+3i)z + C_2\exp((1-3i)z)$

41. $\displaystyle y=C_1\exp(2iz)+C_2\exp(-2iz)+\frac{z}{4}.$

43. $y=(-5-i)\exp(-2z)+(7+2i)\exp(-z)$

45. $$ \begin{align} y&=\exp(5iz) + \exp(-5iz) \\ &=\cos(5z)+i\sin(5z) + \cos(-5z) + i\sin(-5z) \\ &=\cos(5z)+i\sin(5z) + \cos(5z) - i\sin(5z) \\ &=2\cos(5z) \end{align} $$ 47. $v(x)w''+(2v'(x)+p(x)v(x))w' = 0$

49. $\displaystyle \frac{\sqrt{799}}{40}\approx 0.71$ radians/sec

51. $\displaystyle \frac{2\pi}{\sqrt{799}/40} \approx 8.89$ sec

53. $\displaystyle \frac{20\pi}{3} \approx 20.9$ sec

55. $c \approx 346.1$ g/sec

57.

1. 1/sec
2. sec

59. Any value from 2 seconds to infinity

61. Approximately 1.73 seconds

63. $c \approx 77.0$ g/sec

65. Approximately 0.99 seconds

67. From the previous problem we know successive maxima occur at intervals of $2\pi/\omega$ where the position function is $$x(t)=A\exp\left(-\frac{ct}{2m}\right)\cos(\omega t+\phi)$$ for some constants $A$ and $\phi$ (and $ \omega=\sqrt{\frac{k}{m}-\left(\frac{c}{2m}\right)^2},$ but all that matters in what follows is that the $\omega$ in the interval between successive maxima is the same as the omega in the position function). Now $$ \begin{align} x(t+2\pi/\omega)&=A\exp\left(-\frac{c(t+2\pi/\omega)}{2m}\right) \cos(\omega(t+2\pi/\omega)+\phi) \\ &=A\exp\left(-\frac{ct}{2m}\right)\exp\left(-\frac{c\pi}{m}\right) \cos(\omega t + 2\pi + \phi) \\ &=A\exp\left(-\frac{ct}{2m}\right)\exp\left(-\frac{c\pi}{m}\right) \cos(\omega t + \phi) \end{align} $$ Then the ratio of the amplitudes of successive maxima is $$ \begin{align} \frac{x(t+2\pi/\omega)}{x(t)}&= \frac{=A\exp\left(-\frac{ct}{2m}\right)\exp\left(-\frac{c\pi}{m}\right) \cos(\omega t + \phi)} {=A\exp\left(-\frac{ct}{2m}\right)\cos(\omega t + \phi)} \\ &= \exp\left(-\frac{c\pi}{m}\right) \end{align} $$ which is a constant (depending on $c$ and $m$ but not $t$).

69. $4\pi/35 \approx 0.36$ seconds

71. $8\pi^2 kg/sec^2$

73. Approximately $1.39\times10^{-12}$ farads.

75. Coming soon

77. $x(t) = \displaystyle 2B\sin\left(\frac{\omega+\omega_0}{2}t\right) \sin\left(\frac{\omega-\omega_0}{2}t\right)$

79. $y(x)=x\sin(x)+\cos(x)\ln|\cos(x)| + C_1\cos(x)+C_2\sin(x)$

81. $y(x)=2x^2-3x+2x^{-1}$