More complicated ways of solving a problem Feynman says the following,

we shall first study the equation,
  $$d^2x/dt^2 = -x$$
  [...] We assume that in the meantime the Mathematics Department has brought forth a function
  which, when differentiated twice, is equal to itself with a minus sign. (There are, of
  course, ways of getting at this function in a direct fashion, but they are more complicated > than already knowing what the answer is.) The function is $x= \cos t$.

What are these more complicated fashions and where can I learn more about them?
 A: 
What are these more complicated fashions?

Incomplete list, partly based on comments:


*

*Solve $x''+x=0$ by forming the characteristic equation $\lambda^2+1=0$, finding $\lambda=\pm i$, then massaging $x=c_1 e^{i t}+c_2e^{-it}$ into the more digestable form $x=a\cos t+b\sin t$

*Laplace-transform $x''+x=0$ into $s^2 X+X = as+b$ where $a,b$ come from initial conditions at $t=0$. Solve: $X=\dfrac{as}{s^2+1}+\dfrac{b}{s^2+1}$. Invert the Laplace transform:  $x=a\cos t+b\sin t$.

*Look for solution in the form of a power series $x=\sum_{n=0}^\infty a_n t^n$. When plugged into the equation, this will yield $n(n-1)a_n+a_{n-2}=0$ for $n\ge 2$. This will allow you to write down the entire series in terms of the first two coefficiets $a_0,a_1$. (It helps to change the notation by letting $b_n=n!a_n$.) 

*Rewrite the equation in the integral form $x(t)=a+bt-\int_0^t (t-s)x(s)\,ds$, where $a,b$ correspond to the initial conditions. Then run the Picard iteration starting from $x_0(t)=0$. This gives $x_1(t)=a+bt$, $x_2(t)=a+bt-\frac{a}{2}t^2-\frac{b}{6}t^3$, and so on... the result will be the same power series as in method 3.  



and where can I learn more about them? 

In any book on ordinary differential equations, such as the free textbook by  Jiří Lebl. 
