Answer:
A) Particular solution:
![2x+\frac{1}{2}e^{-x}-\frac{8}{5}](https://tex.z-dn.net/?f=2x%2B%5Cfrac%7B1%7D%7B2%7De%5E%7B-x%7D-%5Cfrac%7B8%7D%7B5%7D)
B) Homogeneous solution:
![y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))](https://tex.z-dn.net/?f=y_%7Bh%7D%3De%5E%7B-2x%7D%28c_%7B1%7Dcos%28x%29%2Bc_%7B2%7Dsin%28x%29%29)
C) The most general solution is
![y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}](https://tex.z-dn.net/?f=y%3De%5E%7B-2x%7D%28c_%7B1%7Dcos%28x%29%2Bc_%7B2%7Dsin%28x%29%29%2B2x%2B%5Cfrac%7B1%7D%7B2%7De%5E%7B-x%7D-%5Cfrac%7B8%7D%7B5%7D)
Step-by-step explanation:
Given non homogeneous ODE is
![y''+4y'+5y=10x+e^{-x}---(1)](https://tex.z-dn.net/?f=y%27%27%2B4y%27%2B5y%3D10x%2Be%5E%7B-x%7D---%281%29)
To find homogeneous solution:
![D^{2}+4D+5=0\\D^{2}+4D+4-4+5=0\\\\(D+2)^{2}=-1\\D+2=\pm iD=-2 \pm i\\y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))---(2)](https://tex.z-dn.net/?f=D%5E%7B2%7D%2B4D%2B5%3D0%5C%5CD%5E%7B2%7D%2B4D%2B4-4%2B5%3D0%5C%5C%5C%5C%28D%2B2%29%5E%7B2%7D%3D-1%5C%5CD%2B2%3D%5Cpm%20iD%3D-2%20%5Cpm%20i%5C%5Cy_%7Bh%7D%3De%5E%7B-2x%7D%28c_%7B1%7Dcos%28x%29%2Bc_%7B2%7Dsin%28x%29%29---%282%29)
To find particular solution:
![y_{p}=Ax+B+Ce^{-x}\\\\y'_{p}=A-Ce^{-x}\\y''_{p}=Ce^{-x}\\](https://tex.z-dn.net/?f=y_%7Bp%7D%3DAx%2BB%2BCe%5E%7B-x%7D%5C%5C%5C%5Cy%27_%7Bp%7D%3DA-Ce%5E%7B-x%7D%5C%5Cy%27%27_%7Bp%7D%3DCe%5E%7B-x%7D%5C%5C)
Substituting
in (1)
![y''_{p}+4y'_{p}+5y_{p}=10x+e^{-x}\\Ce^{-x}+4(A-Ce^{-x})+5(Ax+B+Ce^{-x})=10x+e^{-x}\\](https://tex.z-dn.net/?f=y%27%27_%7Bp%7D%2B4y%27_%7Bp%7D%2B5y_%7Bp%7D%3D10x%2Be%5E%7B-x%7D%5C%5CCe%5E%7B-x%7D%2B4%28A-Ce%5E%7B-x%7D%29%2B5%28Ax%2BB%2BCe%5E%7B-x%7D%29%3D10x%2Be%5E%7B-x%7D%5C%5C)
Equating the coefficients
![5Ax+2Ce^{-x}+4A+5B=10x+e^{-x}\\5A=10\\A=2\\4A+5B=0\\B=-\frac{4A}{5}B=-\frac{8}{5}2C=1\\C=\frac{1}{2}\\So,\\y_{p}=2x+\frac{1}{2}e^{-x}-\frac{8}{5}---(3)\\](https://tex.z-dn.net/?f=5Ax%2B2Ce%5E%7B-x%7D%2B4A%2B5B%3D10x%2Be%5E%7B-x%7D%5C%5C5A%3D10%5C%5CA%3D2%5C%5C4A%2B5B%3D0%5C%5CB%3D-%5Cfrac%7B4A%7D%7B5%7DB%3D-%5Cfrac%7B8%7D%7B5%7D2C%3D1%5C%5CC%3D%5Cfrac%7B1%7D%7B2%7D%5C%5CSo%2C%5C%5Cy_%7Bp%7D%3D2x%2B%5Cfrac%7B1%7D%7B2%7De%5E%7B-x%7D-%5Cfrac%7B8%7D%7B5%7D---%283%29%5C%5C)
The general solution is
![y=y_{h}+y_{p}](https://tex.z-dn.net/?f=y%3Dy_%7Bh%7D%2By_%7Bp%7D)
from (2) ad (3)
![y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}](https://tex.z-dn.net/?f=y%3De%5E%7B-2x%7D%28c_%7B1%7Dcos%28x%29%2Bc_%7B2%7Dsin%28x%29%29%2B2x%2B%5Cfrac%7B1%7D%7B2%7De%5E%7B-x%7D-%5Cfrac%7B8%7D%7B5%7D)
A.
6/9 + 6/9 = 12/9
Hope this helps!!!
The cook do not have enough flour. He only has 2.67 cups of flour while the recipe needs 2.75 cups of flour. Therefore, he needs 0.08 more cups of flour in order to have enough for the recipe. Hope this answers the question. Have a nice day.
Answer:
A
Step-by-step explanation:
An educated guess because I really forgot how to do these