Answer:
Therefore the auxiliary solution is ![y=A e^{5x}+Be^{-4x}](https://tex.z-dn.net/?f=y%3DA%20e%5E%7B5x%7D%2BBe%5E%7B-4x%7D)
Therefore
are linearly independent
Step-by-step explanation:
Given, the differential equation is
y"-y'-20 y=0
Let
be the solution of the above differential equation.
y'=
and ![y"= m^2e^{mx}](https://tex.z-dn.net/?f=y%22%3D%20m%5E2e%5E%7Bmx%7D)
Then the above differential equation becomes
![m^2e^{mx}-me^{mx}-20 e^{mx}=0](https://tex.z-dn.net/?f=m%5E2e%5E%7Bmx%7D-me%5E%7Bmx%7D-20%20e%5E%7Bmx%7D%3D0)
![\Rightarrow e^{mx}(m^2-m-20)=0](https://tex.z-dn.net/?f=%5CRightarrow%20e%5E%7Bmx%7D%28m%5E2-m-20%29%3D0)
![\Rightarrow (m^2-m-20)=0](https://tex.z-dn.net/?f=%5CRightarrow%20%28m%5E2-m-20%29%3D0)
![\Rightarrow m^2-5m+4m-20=0](https://tex.z-dn.net/?f=%5CRightarrow%20m%5E2-5m%2B4m-20%3D0)
![\Rightarrow m(m-5) +4(m-5)=0](https://tex.z-dn.net/?f=%5CRightarrow%20m%28m-5%29%20%2B4%28m-5%29%3D0)
![\Rightarrow (m-5)(m+4)=0](https://tex.z-dn.net/?f=%5CRightarrow%20%28m-5%29%28m%2B4%29%3D0)
![\Rightarrow m=5,-4](https://tex.z-dn.net/?f=%5CRightarrow%20m%3D5%2C-4)
If two roots of m are real and distinct then the auxiliary solution is
[where a and b are two roots of m]
Therefore the auxiliary solution is ![y=A e^{5x}+Be^{-4x}](https://tex.z-dn.net/?f=y%3DA%20e%5E%7B5x%7D%2BBe%5E%7B-4x%7D)
Wronskian
![W(e^{-4x},e^{5x})=\left[\begin{array}{cc}e^{-4x}&e^{5x}\\-4e^{-4x}&5e^{5x}\end{array}\right]](https://tex.z-dn.net/?f=W%28e%5E%7B-4x%7D%2Ce%5E%7B5x%7D%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7De%5E%7B-4x%7D%26e%5E%7B5x%7D%5C%5C-4e%5E%7B-4x%7D%265e%5E%7B5x%7D%5Cend%7Barray%7D%5Cright%5D)
![=5e^{-4x}e^{5x}-e^{5x}(-4e^{-4x})](https://tex.z-dn.net/?f=%3D5e%5E%7B-4x%7De%5E%7B5x%7D-e%5E%7B5x%7D%28-4e%5E%7B-4x%7D%29)
≠0
Therefore
are linearly independent.[ ∵W≠0]
He could add / subtract by the past amount with the next amount to figure out.
22.75-15.80= 6.95
15.80-12.83= 2.97
30.50-12.83= 17.67
30.50-7.02= 23.48
Kind of like that, hope that helps! :)