Answer:
Y= 2e^(5t)
Step-by-step explanation:
Taking Laplace of the given differential equation:
s^2+3s-10=0
s^2+5s-2s-10=0
s(s+5)-2(s+5) =0
(s-2) (s+5) =0
s=2, s=-5
Hence, the general solution will be:
Y=Ae^(-2t)+ Be^(5t)………………………………(D)
Put t = 0 in equation (D)
Y (0) =A+B
2 =A+B……………………………………… (i)
Now take derivative of (D) with respect to "t", we get:
Y=-2Ae^(-2t)+5Be^(5t) ....................... (E)
Put t = 0 in equation (E) we get:
Y’ (0) = -2A+5B
10 = -2A+5B ……………………………………(ii)
2(i) + (ii) =>
2A+2B=4 .....................(iii)
-2A+5B=10 .................(iv)
Solving (iii) and (iv)
7B=14
B=2
Now put B=2 in (i)
A=2-2
A=0
By putting the values of A and B in equation (D)
Y= 2e^(5t)
The slope of the trend line is -10
It’s equal to 1/2 in simplest form
Answer:
sad:times
Step-by-step explanation:
