I think it is number 2, x^2-x+2
The correct question is:
Suppose x = c1e^(-t) + c2e^(3t) a solution to x''- 2x - 3x = 0 by substituting it into the differential equation. (Enter the terms in the order given. Enter c1 as c1 and c2 as c2.)
Answer:
x = c1e^(-t) + c2e^(3t)
is a solution to the differential equation
x''- 2x' - 3x = 0
Step-by-step explanation:
We need to verify that
x = c1e^(-t) + c2e^(3t)
is a solution to the differential equation
x''- 2x' - 3x = 0
We differentiate
x = c1e^(-t) + c2e^(3t)
twice in succession, and substitute the values of x, x', and x'' into the differential equation
x''- 2x' - 3x = 0
and see if it is satisfied.
Let us do that.
x = c1e^(-t) + c2e^(3t)
x' = -c1e^(-t) + 3c2e^(3t)
x'' = c1e^(-t) + 9c2e^(3t)
Now,
x''- 2x' - 3x = [c1e^(-t) + 9c2e^(3t)] - 2[-c1e^(-t) + 3c2e^(3t)] - 3[c1e^(-t) + c2e^(3t)]
= (1 + 2 - 3)c1e^(-t) + (9 - 6 - 3)c2e^(3t)
= 0
Therefore, the differential equation is satisfied, and hence, x is a solution.
Answer:
a. 56
b. 72
c. 64
Step-by-step explanation:
Answer: 26
Step-by-step explanation:
divide 19.5 by 3/4 (0.75)
4|y-9| > 36
Two equations:
4(y-9)>36 then y-9>8 then y>17
4(y-9)<-36 then y-9<-8 then y<1
So, the solution is y<1 or y>17
0.5|x+2| > 6
Two equations:
0.5(x+2) > 6 then 0.5x+1>6 then x>10
0.5(x+2) < -6 then 0.5x+1<-6 then x<-14
So, the solution is: x<-14 or x>10
