I am sorry but I do not know but if someone else know help
Answer:
Step-by-step explanation:
Given the differential equation
25y′′+40y′+16y=0
Using D operator to find the complementary solution
Since the Differential equation is equal to 0, then it doesn't have a partial solution.
25y′′+40y′+16y=0
25D² + 40D +16 = 0
25D² + 20D + 20D +16 = 0
5D(5D+4)+4(5D+4)= 0
(5D+4)=0 twice
D = -4/5 twice,
So the solution is a real and equal roots
y(t) = (A+B•t) exp(—0.8t)
Where A and B are constant
The initial value are.
y(0)=a, y′(0)=−1
y(0) = a = (A+B(0)) exp(—0.8(0))
a = A
Then, the constant A = a.
Now, y'(t)
y'(t)=B•exp(-0.8t)-0.8(A+B•t)exp(-0.8t
-1 = B - 0.8A
Since A =a
Then, B = 0.8a —1
The solution becomes
y(t) = (a+(0.8a-1)•t) exp(—0.8t)
For t>0
For positive,
a + (0.8a -1)> 0
1.8a > 1
a > 1/1.8
a > 10/18 > 5/9
a > 5/9.
We just substitute in -12 for x in each one and see if it's true
2(-12)+6 = -24 + 6 = -18, not 30, NOPE
-0.25(-12) - 1 = 3-1 = 2, that worked, CHECK ME
-12 - 8 = -20, CHECK ME
-35(-12) +2 = 422, not -515, NOPE
3(-12+25)=3(13)=39, CHECK ME
The correct answer is 10 \8.
