The 2 in the hundreds place is 1000* the 2 in the tenths place
I'll assume the ODE is
![y'' - 3y' + 2y = e^x + e^{2x} + e^{-x}](https://tex.z-dn.net/?f=y%27%27%20-%203y%27%20%2B%202y%20%3D%20e%5Ex%20%2B%20e%5E%7B2x%7D%20%2B%20e%5E%7B-x%7D)
Solve the homogeneous ODE,
![y'' - 3y' + 2y = 0](https://tex.z-dn.net/?f=y%27%27%20-%203y%27%20%2B%202y%20%3D%200)
The characteristic equation
![r^2 - 3r + 2 = (r - 1) (r - 2) = 0](https://tex.z-dn.net/?f=r%5E2%20-%203r%20%2B%202%20%3D%20%28r%20-%201%29%20%28r%20-%202%29%20%3D%200)
has roots at
and
. Then the characteristic solution is
![y = C_1 e^x + C_2 e^{2x}](https://tex.z-dn.net/?f=y%20%3D%20C_1%20e%5Ex%20%2B%20C_2%20e%5E%7B2x%7D)
For nonhomogeneous ODE (1),
![y'' - 3y' + 2y = e^x](https://tex.z-dn.net/?f=y%27%27%20-%203y%27%20%2B%202y%20%3D%20e%5Ex)
consider the ansatz particular solution
![y = axe^x \implies y' = a(x+1) e^x \implies y'' = a(x+2) e^x](https://tex.z-dn.net/?f=y%20%3D%20axe%5Ex%20%5Cimplies%20y%27%20%3D%20a%28x%2B1%29%20e%5Ex%20%5Cimplies%20y%27%27%20%3D%20a%28x%2B2%29%20e%5Ex)
Substituting this into (1) gives
![a(x+2) e^x - 3 a (x+1) e^x + 2ax e^x = e^x \implies a = -1](https://tex.z-dn.net/?f=a%28x%2B2%29%20e%5Ex%20-%203%20a%20%28x%2B1%29%20e%5Ex%20%2B%202ax%20e%5Ex%20%3D%20e%5Ex%20%5Cimplies%20a%20%3D%20-1)
For the nonhomogeneous ODE (2),
![y'' - 3y' + 2y = e^{2x}](https://tex.z-dn.net/?f=y%27%27%20-%203y%27%20%2B%202y%20%3D%20e%5E%7B2x%7D)
take the ansatz
![y = bxe^{2x} \implies y' = b(2x+1) e^{2x} \implies y'' = b(4x+4) e^{2x}](https://tex.z-dn.net/?f=y%20%3D%20bxe%5E%7B2x%7D%20%5Cimplies%20y%27%20%3D%20b%282x%2B1%29%20e%5E%7B2x%7D%20%5Cimplies%20y%27%27%20%3D%20b%284x%2B4%29%20e%5E%7B2x%7D)
Substitute (2) into the ODE to get
![b(4x+4) e^{2x} - 3b(2x+1)e^{2x} + 2bxe^{2x} = e^{2x} \implies b=1](https://tex.z-dn.net/?f=b%284x%2B4%29%20e%5E%7B2x%7D%20-%203b%282x%2B1%29e%5E%7B2x%7D%20%2B%202bxe%5E%7B2x%7D%20%3D%20e%5E%7B2x%7D%20%5Cimplies%20b%3D1)
Lastly, for the nonhomogeneous ODE (3)
![y'' - 3y' + 2y = e^{-x}](https://tex.z-dn.net/?f=y%27%27%20-%203y%27%20%2B%202y%20%3D%20e%5E%7B-x%7D)
take the ansatz
![y = ce^{-x} \implies y' = -ce^{-x} \implies y'' = ce^{-x}](https://tex.z-dn.net/?f=y%20%3D%20ce%5E%7B-x%7D%20%5Cimplies%20y%27%20%3D%20-ce%5E%7B-x%7D%20%5Cimplies%20y%27%27%20%3D%20ce%5E%7B-x%7D)
and solve for
.
![ce^{-x} + 3ce^{-x} + 2ce^{-x} = e^{-x} \implies c = \dfrac16](https://tex.z-dn.net/?f=ce%5E%7B-x%7D%20%2B%203ce%5E%7B-x%7D%20%2B%202ce%5E%7B-x%7D%20%3D%20e%5E%7B-x%7D%20%5Cimplies%20c%20%3D%20%5Cdfrac16)
Then the general solution to the ODE is
![\boxed{y = C_1 e^x + C_2 e^{2x} - xe^x + xe^{2x} + \dfrac16 e^{-x}}](https://tex.z-dn.net/?f=%5Cboxed%7By%20%3D%20C_1%20e%5Ex%20%2B%20C_2%20e%5E%7B2x%7D%20-%20xe%5Ex%20%2B%20xe%5E%7B2x%7D%20%2B%20%5Cdfrac16%20e%5E%7B-x%7D%7D)
Answer:
A) But if anyone thinks this they should go back to preschool
Step-by-step explanation:
On any other value of x you would take the numerators and denominators and multiply them across:
![\frac{x}{3}*\frac{6}{x}=\frac{x*6}{x*3}=\frac{6}{3} =2](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7B3%7D%2A%5Cfrac%7B6%7D%7Bx%7D%3D%5Cfrac%7Bx%2A6%7D%7Bx%2A3%7D%3D%5Cfrac%7B6%7D%7B3%7D%20%3D2)
But, this literally does not make since considering the fact that you cannot divide any value by zero. Try it, it won't work. If we plug these into a trusty dusty calculator, we get this:
![\frac{0}{3}=0\\\\\frac{6}{0}=DOMAINERROR](https://tex.z-dn.net/?f=%5Cfrac%7B0%7D%7B3%7D%3D0%5C%5C%5C%5C%5Cfrac%7B6%7D%7B0%7D%3DDOMAINERROR)
0*DOMAINERROR= You can't do it.
There are other reasons why you can't divide by zero, I would suggest looking them up on YT and learning about them.
I would answer A, but the person who wrote this question just wasn't focusing when they wrote the question.
J can find the height of the fifth member on this way
S=150
S1=153
S2=150
S3=151
S4=152
S5=?
S= (s1+s2+s3+s4+s5)/5:
5*150= <span>153+150+151+152+S5
S5=5*150-606=750-606=144
Answer is 144 cm</span>