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)
Simple.
to figure out if they are either: equal, greater, or smaller...
First, find a common denominator...
![\frac{21}{30}](https://tex.z-dn.net/?f=%20%5Cfrac%7B21%7D%7B30%7D%20)
and
![\frac{2}{3} * \frac{10}{10} = \frac{20}{30}](https://tex.z-dn.net/?f=%20%5Cfrac%7B2%7D%7B3%7D%20%2A%20%5Cfrac%7B10%7D%7B10%7D%20%3D%20%5Cfrac%7B20%7D%7B30%7D%20)
Now, we can
obviously make the assumption that
![\frac{21}{30}](https://tex.z-dn.net/?f=%20%5Cfrac%7B21%7D%7B30%7D%20)
is greater than
![\frac{20}{30}](https://tex.z-dn.net/?f=%20%5Cfrac%7B20%7D%7B30%7D%20)
.
Thus, your answer.
Six million seven thousand two hundred
Step-by-step explanation:
It's 6r+5y. But it's not in the option
Answer:
7/3
Explanation:
The slope of the line is equal to the rise/run, or in other words, the number of units the line travels upwards over the number of units the line travels to the right.
We can identify the slope using any two points on the line. Here, we can use the two points that are marked on the picture. The second point is 7 units above the first and 3 units to the right of the first, so the slope of the line is equal to 7/3.
Another way to calculate the slope of the line is the use this formula:
(y2-y1)/(x2-x1)
The first point is at the coordinate (1,-4) and second point is at the coordinate (4,3). When we plug these two coordinates into the equation, we get this:
(y2-y1)/(x2-x1)
->(3-(-4))/(4-1)
When we simply the fraction, we would get 7/3 and that would give us the slope.
I hope this helps!