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 (0, infinity).
To verify that y = e^|x| is a solution to the given differential equation on the intervals (-∞, 0) and (0, ∞), we need to substitute it into the differential equation and check if it satisfies the equation.
Let's start with the interval (-∞, 0):
For x < 0, the given differential equation is y' - y = -2e^(-x).
Since y = e^|x|, we find y' by taking the derivative of y with respect to x:
dy/dx = d/dx(e^|x|).
Let's break this down into two cases:
1. For x < 0, the absolute value |x| is equal to -x. So, e^|x| becomes e^(-x).
dy/dx = d/dx(e^(-x)).
Differentiating e^(-x) with respect to x, using the chain rule, we get:
dy/dx = -e^(-x).
Now, let's substitute these values into the differential equation:
-2e^(-x) - e^|x| = -2e^(-x) - e^(-x).
Simplifying the right side, we get:
-2e^(-x) - e^(-x) = -3e^(-x).
So, the differential equation becomes:
-3e^(-x) = -3e^(-x).
Since the left side is equal to the right side of the equation, y = e^|x| satisfies the differential equation on the interval (-∞, 0).
Now, let's move on to the interval (0, ∞):
For x ≥ 0, the given differential equation is y' - y = 0.
Since y = e^|x|, we find y' by taking the derivative of y with respect to x:
dy/dx = d/dx(e^|x|).
Again, let's break this down into two cases:
2. For x ≥ 0, the absolute value |x| is equal to x. So, e^|x| remains as e^|x|.
dy/dx = d/dx(e^|x|).
Differentiating e^|x| with respect to x, using the chain rule, we get:
dy/dx = e^|x| * d/dx(|x|).
The derivative of |x| with respect to x is the sign function, which equals 1 for x > 0 and 0 for x = 0.
Therefore, dy/dx = e^|x| * 1 = e^|x|.
Now, let's substitute these values into the differential equation:
0 - e^|x| = -e^|x|.
Since the left side is equal to the right side of the equation, y = e^|x| satisfies the differential equation on the interval (0, ∞).
Therefore, we have verified that y = e^|x| is a solution to the given differential equation on each of the intervals (-∞, 0) and (0, ∞).