Step-by-step explanation:
Let y1 and y2 be (e^x)/2, and (xe^x)/2 respectively.
The Wronskian of them functions be
W = (y1y2' - y1'y2)
y1 = (e^x)/2 = y1'
y2 = (xe^x)/2
y2' = (1/2)(x + 1)e^x
W = (1/4)(x + 1)e^(2x) - (1/4)xe^(2x)
= (1/4)(x + 1 - x)e^(2x)
W = (1/4)e^(2x)
Since the Wronskian ≠ 0, we conclude that functions are linearly independent, and hence, form a set of fundamental solutions.
Answer:
Correct option: D
Step-by-step explanation:
If we want to shift the graph one unit to the left, we need to add one in the argument of the function, that is, we need to find G(x+1), and we do that by switching the values of x by (x+1) in the equation G(x)
So we have that:
G(x) = x^3 - x^2
Switching all values of x by x+1, we have:
G(x+1) = (x+1)^3 - (x+1)^2
So the correct option is D.
(Y1-Y2)/(X1-X2)
(2-4)/(-2-3)
-2/-5
2/5
