3. $\displaystyle \sum_{n=0}^{n=\infty} \frac{x^{2n}}{(2n)!} $

5. $\displaystyle \sum_{n=1}^{n=\infty} \frac{2^{n-1}x^n}{n!} $

7. We are given $\displaystyle y(x) = \sum_{n=0}^{\infty} a_{2n}x^{2n} $ since the coefficients of all the odd terms are 0. Then $$ y(-x) = \sum_{n=0}^{\infty} a_{2n}(-x)^{2n} = \sum_{n=0}^{\infty} a_{2n}x^{2n} = y(x) $$ and so $y(x)$ is an even function.

9. The recurrence relations are $$ \begin{align} a_2 &= -\frac{a_0}{2} \\ a_3 &= -\frac{3a_1}{6} \\ a_{n+2} &= -\frac{(2n+1)a_n + a_{n-2}}{(n+2)(n+1)},\qquad (n \ge 2) \end{align} $$ Since $a_1 = y'(0)= 0,$ it follows from the second equation that $a_3=0,$ and then since each odd coefficient depends only on the two previous odd coefficients (i.e. $a_{n+2}$ depends only on $a_n$ and $a_{n-2}$ and if $n+2$ is odd then so are $n$ and $n-2$), it follows that all odd coefficients will be 0. Hence the function is even by problem 7.

11. Since $a_0\ne 0$ and $a_1\ne 0,$ it follows that there are non-zero coefficents among both the odd and even terms and so the function is neither odd nor even. Now in problems 7 and 8, we only showed that if the odd or even terms vanish, then the function is even or odd. We did not show that if the function is even or odd, then the odd or even terms must necessarily vanish. However, the converse is indeed true.

13. $\displaystyle -\sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n-1)(2n+1)}$ or, equivalently, $\displaystyle \sum_{n=0}^{\infty} \frac{x^{2n+1}}{1-4n^2}$

15. $\displaystyle a_{n+2} = -\frac{n^2-n+1}{4n^2+12n+8}a_n$

17. Assuming $n$ is even (since all the odd coefficients are 0), we have $$ \begin{align} \lim_{n\to\infty}\left| \frac{a_{n+2}x^{n+2}}{a_nx^n}\right| &= \lim_{n\to\infty}\left| \frac{a_{n+2}}{a_n}\right| x^2 \\ &= \lim_{n\to\infty}\left| \frac{n^2-n+1}{4n^2+12n+8}\right|x^2 \\ &= \lim_{n\to\infty}\left| \frac{1-1/n+1/n^2}{4+12/n+8/n^2}\right|x^2 \\ &= \frac{x^2}{4} \end{align} $$ So $\displaystyle \lim_{n\to\infty}\left| \frac{a_{n+2}x^{n+2}}{a_nx^n}\right| < 1$ if $\displaystyle \frac{x^2}{4} < 1,$ or $|x| < 2,$ while $\displaystyle \lim_{n\to\infty}\left| \frac{a_{n+2}x^{n+2}}{a_nx^n}\right| > 1$ if $\displaystyle \frac{x^2}{4} > 1,$ or $|x| > 2.$ So the series converges for $|x| < 2$ and diverges for $|x| > 2,$ hence the radius of convergence is 2.

19. $\displaystyle y = c_1x^{-1} + c_2x^{-3}.$

21. $\displaystyle y = c_1\sqrt{x} + c_2x^3.$

23. $\displaystyle y=\frac{x^4}{2} + c_1x^2 + c_2x^3.$

25. $\displaystyle y = x\exp(x)+c_1x + c_2x^2. $

27. $\displaystyle y = c_1\cos(\log|x^2|) + c_2\sin(\log|x^2|). $

29. $\displaystyle y = c_1x^2\cos(\log|x^2|) + c_2x^2\sin(\log|x^2|). $

31. $\displaystyle y = \frac{c_1}{(x+2)^2} + \frac{c_2}{(x+2)^4}. $

33. -1 is regular, -2 is regular.

35. 0 is irregular.

37. -1 is regular, 1 is regular.

39. $2\pi n$ is regular for all integer $n$.

41. $0 < |x| < \infty$

43. $0 < |x-1| < 2$

45. -1 and -3

47. $1 \pm i\sqrt{3}$

49. $-1 \pm \sqrt{11}$

51. $$\frac{\partial u}{\partial x} = \sin(\phi)\cos(\theta)\frac{\partial u}{\partial \rho} - \frac{\cos(\theta)}{\rho \sin(\phi)} \frac{\partial u}{\partial \theta} + \frac{\cos(\theta)}{\rho} \frac{\partial u}{\partial \phi} $$ 53. $$ \frac{\partial u}{\partial z} = \cos(\phi)\frac{\partial u}{\partial \rho} - \frac{\sin(\phi)}{\rho}\frac{\partial u}{\partial \phi} $$ 55. $$ \begin{align} \frac{\partial^2 u}{\partial y^2} = &\sin^2(\theta)\sin^2(\phi)\frac{\partial^2 u}{\partial \rho^2} + \frac{\cos^2(\theta)}{\rho^2\sin^2(\phi)}\frac{\partial^2 u}{\partial \theta^2} + \frac{\sin^2(\theta)\cos^2(\phi)}{\rho^2}\frac{\partial^2 u}{\partial \phi^2} \\ &+ \frac{2\sin(\theta)\cos(\theta)}{\rho}\frac{\partial^2 u}{\partial \rho \partial \theta} + \frac{2\sin^2(\theta)\sin(\phi)\cos(\phi)}{\rho} \frac{\partial^2 u}{\partial \rho \partial \phi} + \frac{2\sin(\theta)\cos(\theta)\cos(\phi)}{\rho^2\sin(\phi)} \frac{\partial^2 u}{\partial \theta \partial \phi} \\ &+ \frac{\cos^2(\theta)+\sin^2(\theta)\cos^2(\phi)}{\rho}\frac{\partial u}{\partial \rho} - \frac{2\sin(\theta)\cos(\theta)}{\rho^2\sin^2(\phi)}\frac{\partial u}{\partial \theta} + \frac{\cos^2(\theta)\cos(\phi)-2\sin^2(\theta)\sin^2(\phi)\cos(\phi)}{\rho^2 \sin(\phi)}\frac{\partial u}{\partial \phi} \end{align} $$ 57. $$ \frac{\partial^2 u}{\partial \rho^2} + \frac{2}{\rho}\frac{\partial u}{\partial \rho} + \frac{1}{\rho^2\sin(\phi)} \frac{\partial^2 u}{\partial \theta^2} + \frac{1}{\rho^2}\frac{\partial^2 u}{\partial \phi^2} + \frac{\cos(\phi)}{\rho^2 \sin(\phi)}\frac{\partial u}{\partial \phi} =0 $$

59. $\displaystyle y(x)=(2e^2)x\exp(-2x) = 2x\exp(2-2x)$

61. $\displaystyle y(x) = \exp(-x)\cos(2x) + \frac{2e-\cos(2)}{\sin(2)}\exp(-x)\sin(2x) $

63. $\displaystyle x_1 = \frac{n\pi}{2}$ for any integer $n$

65. $\displaystyle \alpha = (n\pi)^2$ for any integer $n$

67. $\displaystyle \alpha = 4+(n\pi)^2$ for any integer $n$

69. $$ \begin{align} T''(t)X(x) &= cT(t)X''(x) \\ \frac{T''(t)X(x)}{T(t)X(x)} & = \frac{cT(t)X''(x)}{T(t)X(x)} \\ \frac{T''(t)}{T(t)}&=\frac{cX''(x)}{X(x)} \end{align}$$ Now since the only way a function of $t$ alone can equal a function of $x$ alone is if they are constant. So we have $$ \frac{T''(t)}{T(t)}=\frac{cX''(x)}{X(x)}=\lambda $$ and so we have the two ordinary differential equations $$ \begin{align} T''(t)-\lambda T(t) &= 0 \\ cX''(x)-\lambda X(x) &=0 \end{align} $$