Verify that y=e^abs(x) is a solution to
y'-y={-2e^-x, x<0
{0, x>= 0
on each of the intervals (-infinity,0) and (o, infinity).
To verify that y = e^|x| is a solution to the given differential equation, we need to substitute y into the equation and check if it satisfies the equation on each interval.
Let's start by differentiating y = e^|x| to find y'.
On the interval (-∞, 0):
For x < 0, y = e^(-x), so let's find y':
y = e^(-x)
Differentiating both sides with respect to x:
y' = (-e^(-x)) * (-1) (by the chain rule)
y' = e^(-x)
Now, substitute y' and y into the differential equation: y' - y = -2e^(-x)
e^(-x) - e^(-x) = -2e^(-x)
0 = -2e^(-x)
Since the left-hand side (LHS) is equal to zero and the right-hand side (RHS) is equal to -2e^(-x) for x < 0, the differential equation is satisfied on the interval (-∞, 0) when y = e^(-x).
Now, let's check the interval (0, ∞):
For x ≥ 0, y = e^(x), so let's find y':
y = e^(x)
Differentiating both sides with respect to x:
y' = e^(x)
Now, substitute y' and y into the differential equation: y' - y = 0
e^(x) - e^(x) = 0
0 = 0
Both the LHS and the RHS are equal to zero for x ≥ 0, so the differential equation is satisfied on the interval (0, ∞) when y = e^(x).
Therefore, y = e^|x| is a solution to the given differential equation on each of the intervals (-∞,0) and (0,∞).