Answer:
![y(t)\ =\ C_1e^{-2t}+C_2e^t\ -\ \dfrac{1}{2}](https://tex.z-dn.net/?f=y%28t%29%5C%20%3D%5C%20C_1e%5E%7B-2t%7D%2BC_2e%5Et%5C%20-%5C%20%5Cdfrac%7B1%7D%7B2%7D)
Step-by-step explanation:
Given differential equation is
y"+y'-2y = 1
![=>\ (D^2\ +\ D\ -\ 2)y\ =\ 1](https://tex.z-dn.net/?f=%3D%3E%5C%20%28D%5E2%5C%20%2B%5C%20D%5C%20-%5C%202%29y%5C%20%3D%5C%201)
Hence, the characteristics is
![D^2+D-2\ =\ 0](https://tex.z-dn.net/?f=D%5E2%2BD-2%5C%20%3D%5C%200)
![=>\ D^2\ +\ 2D\ -\ D\ -2\ =\ 0](https://tex.z-dn.net/?f=%3D%3E%5C%20D%5E2%5C%20%2B%5C%202D%5C%20-%5C%20D%5C%20-2%5C%20%3D%5C%200)
![=> (D+2)(D-1) = 0](https://tex.z-dn.net/?f=%3D%3E%20%28D%2B2%29%28D-1%29%20%3D%200)
=> D = -2, 1
The general equation of the given differential equation is
![y_c(t)\ =\ C_1e^{-2t}+C_2e^t](https://tex.z-dn.net/?f=y_c%28t%29%5C%20%3D%5C%20C_1e%5E%7B-2t%7D%2BC_2e%5Et)
Let's consider that
![y_2(t)\ =\ e^{t}](https://tex.z-dn.net/?f=y_2%28t%29%5C%20%3D%5C%20e%5E%7Bt%7D)
![y'_2(t)\ =\ e^t](https://tex.z-dn.net/?f=y%27_2%28t%29%5C%20%3D%5C%20e%5Et)
g(t) = 1
Wronskian can be given by,
W = y_1(t)y'_2(t) - y_2(t)y'_1(t)
![=\ e^{-2t}.e^t\ -\ e^{t}.(-2e^{-2t})](https://tex.z-dn.net/?f=%3D%5C%20e%5E%7B-2t%7D.e%5Et%5C%20-%5C%20e%5E%7Bt%7D.%28-2e%5E%7B-2t%7D%29)
![=\ e^{-t}\ +\ 2e^{-t}](https://tex.z-dn.net/?f=%3D%5C%20e%5E%7B-t%7D%5C%20%2B%5C%202e%5E%7B-t%7D)
![=\ 3e^{-t}](https://tex.z-dn.net/?f=%3D%5C%203e%5E%7B-t%7D)
Now, the particular integral can be given by
![y_p(t)\ = \ -y_1(t)\int{\dfrac{y_2(t).g(t)}{W}dt}\ +\ y_2(t)\int{\dfrac{y_1(t).g(t)}{W}dt}](https://tex.z-dn.net/?f=y_p%28t%29%5C%20%3D%20%5C%20-y_1%28t%29%5Cint%7B%5Cdfrac%7By_2%28t%29.g%28t%29%7D%7BW%7Ddt%7D%5C%20%2B%5C%20y_2%28t%29%5Cint%7B%5Cdfrac%7By_1%28t%29.g%28t%29%7D%7BW%7Ddt%7D)
![=\ -e^{-2t}\int{\dfrac{e^t\times 1}{3e^{-t}}dt}\ +\ e^t\int{\dfrac{e^{-2t}\times 1}{3e^{-t}}dt}](https://tex.z-dn.net/?f=%3D%5C%20-e%5E%7B-2t%7D%5Cint%7B%5Cdfrac%7Be%5Et%5Ctimes%201%7D%7B3e%5E%7B-t%7D%7Ddt%7D%5C%20%2B%5C%20e%5Et%5Cint%7B%5Cdfrac%7Be%5E%7B-2t%7D%5Ctimes%201%7D%7B3e%5E%7B-t%7D%7Ddt%7D)
![=\ -e^{-2t}\int{\dfrac{e^{2t}}{3}dt}\ +\ e^t\int{\dfrac{e^{-t}}{3}dt}](https://tex.z-dn.net/?f=%3D%5C%20-e%5E%7B-2t%7D%5Cint%7B%5Cdfrac%7Be%5E%7B2t%7D%7D%7B3%7Ddt%7D%5C%20%2B%5C%20e%5Et%5Cint%7B%5Cdfrac%7Be%5E%7B-t%7D%7D%7B3%7Ddt%7D)
![=\ (-e^{-2t})(\dfrac{e^{2t}}{6})\ +\ (e^t)(\dfrac{e^{-t}}{-3})](https://tex.z-dn.net/?f=%3D%5C%20%28-e%5E%7B-2t%7D%29%28%5Cdfrac%7Be%5E%7B2t%7D%7D%7B6%7D%29%5C%20%2B%5C%20%28e%5Et%29%28%5Cdfrac%7Be%5E%7B-t%7D%7D%7B-3%7D%29)
![=\ \dfrac{-1}{6}\ -\ \dfrac{1}{3}](https://tex.z-dn.net/?f=%3D%5C%20%5Cdfrac%7B-1%7D%7B6%7D%5C%20-%5C%20%5Cdfrac%7B1%7D%7B3%7D)
![=\ \dfrac{-3}{6}](https://tex.z-dn.net/?f=%3D%5C%20%5Cdfrac%7B-3%7D%7B6%7D)
![=\ \dfrac{-1}{2}](https://tex.z-dn.net/?f=%3D%5C%20%5Cdfrac%7B-1%7D%7B2%7D)
Now,
![y(t)\ =\ y_c(t)\ +\ y_p(t)](https://tex.z-dn.net/?f=y%28t%29%5C%20%3D%5C%20y_c%28t%29%5C%20%2B%5C%20y_p%28t%29)
![=\ C_1e^{-2t}+C_2e^t\ -\ \dfrac{1}{2}](https://tex.z-dn.net/?f=%3D%5C%20C_1e%5E%7B-2t%7D%2BC_2e%5Et%5C%20-%5C%20%5Cdfrac%7B1%7D%7B2%7D)
Hence, the complete solution of the given differential equation is
![y(t)\ =\ C_1e^{-2t}+C_2e^t\ -\ \dfrac{1}{2}](https://tex.z-dn.net/?f=y%28t%29%5C%20%3D%5C%20C_1e%5E%7B-2t%7D%2BC_2e%5Et%5C%20-%5C%20%5Cdfrac%7B1%7D%7B2%7D)
<h3>
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➷ volume = length x width x height
volume = 2.5 x 1.5 x 4
volume = 15 in^3
<h3><u>
✽</u></h3>
➶ Hope This Helps You!
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➶ Have A Great Day ^-^
↬ ʜᴀɴɴᴀʜ ♡
Answer:
the average of data is which is found by subtracting third and fourth term
Answer:
9 hours
Step-by-step explanation:
if 8 volunteers=6 hours then
1 volunteer= 6÷8 =0.75 hours
12 volunteers= 0.75×12 =9 hours