First of all in this case the solution to the first order linear differential equation is missing an arbitrary constant. It should say:
y = e-2x ( C + ∫ e-2x b(x) dx )
There are no limits of integration in this formula: the integral is an indefinite one. What has to be done instead is to do the integral for the different cases in the piecewise definition of b(x). This means that the answer for y will also be piecewise.
In the case where |x| > 1, we have b(x) = 0 and thus:
y = e-2x ( C + ∫ e-2x * 0 * dx ) = C e-2x
Here the integral becomes zero and the only part of the solution left is the C e-2x (this is why it's important to include the constant, otherwise one might conclude the only solution was y = 0).
For the case where |x| ≤ 1, we have b(x) = 1 - | x |. We don't want to deal with integrating absolute values, so we split this into two sub-cases. First, for 0 ≤ x ≤ 1, we can replace |x| with x and thus b(x) = 1 - x on this interval. So for 0 ≤ x ≤ 1, we have:
y = e-2x ( C + ∫ e-2x * (1 - x) * dx )
Doing the integral or solving with, say, Wolfram Alpha gives:
∫ e-2x * (1 - x) * dx = (1/4) * e-2x (2x - 1)
And so for 0 ≤ x ≤ 1 we have
y = C e-2x + (1/4)*(2x - 1)
For -1 ≤ x < 0, this time we replace |x| with -x. So b(x) = 1 + x and we have
y = e-2x ( C + ∫ e-2x * (1 + x) * dx )
Solving similarly to the previous integral, we get:
∫ e-2x * (1 + x) * dx = (-1/4) * e-2x (2x + 3)
And so for -1 ≤ x ≤ 0 we have
y = C e-2x - (1/4)*(2x + 3)
So, putting it all together, the solution is:
y(x) =
C e-2x - (1/4)*(2x + 3) for -1 ≤ x ≤ 0
C e-2x + (1/4)*(2x - 1) for 0 ≤ x ≤ 1
C e-2x otherwise
No initial condition is specified and this is a first-order differential equation, so it makes sense that the answer contains a single arbitrary constant.
Hope this helps!
