fix some math

src/8_differential_equations_2.md

@@ -198,7 +198,7 @@ $$f(x) = e^{\lambda_1 x}, \ x e^{\lambda_1 x} , \ \cdots, \ x^{m_{1}-1} e^{\lamb
     since they are linear combinations of $f_1$ and $f_2$ which remain linearly
     independent,
     
-    $$\Tilde{f_1}(x)=cos(kx), \ \Tilde{f_2}(x)=sin{kx}.$$
+    $$\tilde{f_1}(x)=cos(kx), \tilde{f_2}(x)=sin{kx}.$$
     
     **Case 2: $E<0$**
     This time, define $E=-k^2$, for constant $k$. The characteristic polynomial 
@@ -397,9 +397,9 @@ In the previous equation, the coefficient $A$ can be determined if the original
 PDE was supplied with an initial condition. 
 
 Putting the solutions to the two ODEs together and redefining 
-$\Tilde{A}=A \cdot c_1$, we arrive at the solutions for theb PDE,
+$\tilde{A}=A \cdot c_1$, we arrive at the solutions for theb PDE,
 
-$\psi_n(x,t) = \Tilde{A}_n e^{-i \frac{\lambda_n t}{\hbar}} sin(\frac{n \pi x}{L}).$
+$\psi_n(x,t) = \tilde{A}_n e^{-i \frac{\lambda_n t}{\hbar}} sin(\frac{n \pi x}{L}).$
 
 Notice that there is one solution $\psi_{n}(x,t)$ for each natural number $n$. 
 These are still very special solutions. We will begin discussing next how to