Hello,
I am going to remember:
y'+3y=0==>y=C*e^(-3t)
y'=C'*e^(-3t)-3C*e^(-3t)
y'+3y=C'*e^(-3t)-3Ce^(-3t)+3C*e^(-3t)=C'*e^(-3t) = t+e^(-2t)
==>C'=(t+e^(-2t))/e^(-3t)=t*e^(3t)+e^t
==>C=e^t+t*e^(3t) /3-e^(3t)/9
==>y= (e^t+t*e^(3t)/3-e^(3t)/9)*e^(-3t)+D
==>y=e^(-2t)+t/3-1/9+D
==>y=e^(-2t)+t/3+k
Answer:
Quadrant I and III
Step-by-step explanation:
The coordinate (3,9) is all positive, therefore it lies in quadrant I.
The coordinate (-3,-9) is all negative, therefore it lies in quadrant III.
Answer:
3x^2 + 8
Step-by-step explanation:
Given functions,
f(x) = 3x + 8 ----(1)
g(x) = x^2 -----(2)
Since,
(fog)(x) = f( g(x) ) ( Composition of functions )
=f(x^2) ( From equation (2) )
=3x^2 + 8 ( From equation (1) )
Hence,
(fog)(x)=3x^2 + 8
Step-by-step explanation:
