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.
-8 and -9 because it is negative
You need to type the same equation that’s shown above of the text box. If there’s still a technical issue, let me know, and I’ll try to help!
Answer:
26
Step-by-step explanation:
Since the input is -4, x = -4. The equation then is y = -7(-4) -2
A negative times a negative is a positive, so -7 times -4 is 28.
Now the equation reads y = 28 - 2
Then obviously you just subtract the two and then you get 26.
Spacing needs to be more clear instead of having a giant paragraph. you do not write math in giant paragraphs do you