Answer:
g(t) = 0
And
The differential equation 
is linear and homogeneous
Step-by-step explanation:
Given that,
The differential equation is -
![e^{t}y' + (9t - \frac{1}{t^{2} + 81 } )y = 0\\e^{t}y' + (\frac{9t(t^{2} + 81 ) - 1}{t^{2} + 81 } )y = 0\\e^{t}y' + (\frac{9t^{3} + 729t - 1}{t^{2} + 81 } )y = 0\\y' + [\frac{9t^{3} + 729t - 1}{e^{t}(t^{2} + 81) } ]y = 0](https://tex.z-dn.net/?f=e%5E%7Bt%7Dy%27%20%2B%20%289t%20-%20%5Cfrac%7B1%7D%7Bt%5E%7B2%7D%20%2B%2081%20%7D%20%29y%20%3D%200%5C%5Ce%5E%7Bt%7Dy%27%20%2B%20%28%5Cfrac%7B9t%28t%5E%7B2%7D%20%2B%2081%20%29%20-%201%7D%7Bt%5E%7B2%7D%20%2B%2081%20%7D%20%29y%20%3D%200%5C%5Ce%5E%7Bt%7Dy%27%20%2B%20%28%5Cfrac%7B9t%5E%7B3%7D%20%2B%20729t%20%20-%201%7D%7Bt%5E%7B2%7D%20%2B%2081%20%7D%20%29y%20%3D%200%5C%5Cy%27%20%2B%20%5B%5Cfrac%7B9t%5E%7B3%7D%20%2B%20729t%20%20-%201%7D%7Be%5E%7Bt%7D%28t%5E%7B2%7D%20%2B%2081%29%20%7D%20%5Dy%20%3D%200)
By comparing with y′+p(t)y=g(t), we get
g(t) = 0
And
The differential equation is linear and homogeneous.
I thought I knew the answer but I don’t Srry I need points
Answer:
f(2) =-5
Step-by-step explanation:
We want to find the value of the function when x =2
When x=2 the y value is -5
f(2) =-5
5p + 4g = 42.5
<span>3p + 6g = 34.5 </span>
<span>15p + 12g = 127.5 </span>
<span>-6p - 12g = -69 </span>
<span>9p = 58.5 </span>
<span>p = 6.5 </span>
<span>g = 2.5 </span>
<span>popcorn = $6.50 </span>
<span>granola = $2.50
A is the answer </span>
