How to Solve Exact Differential Equations with Discontinuous Functions?

[mehtamonica]

i have to solve the following differential equation :

dy/ dx + y = f(x) where f (x) = 2, 0 <= x < 1 and 0 if x >=1

and y (0)=0.

please explain how to solve it as it involves a discontinuous function ?

I am stuck while computing after computing the integrating factor e^x. Please suggest how to proceed beyond this.

 

[HallsofIvy]

Yes, the integrating factor is e^x. Multiplying both side of the equation by e^x you have
e^x\frac{dy}{dx}+ e^xy= e^xf(x)
\frac{d(e^xy)}{dx}= e^xf(x)
Integrate both sides to get
e^x y(x)= \int_0^x tf(t)dt
(I chose the lower limit of integration to be 0 because the integral of any function from 0 to 0 so that gives y(0)= 0.)

I suspect it is that integration that is bothering you. Take it "step by step".

For 0\le x\le 1 f(x)= 2 so e^xf(x)= 2e^x on that interval.
\int_0^x 2e^t dt= 2(e^x- e^0)= 2(e^x- 1)
Of course, for that, if x= 1, 2(e- 1)

For x> 1, f(x)= 0 so e^x f(x)= 0. \int_1^x 0 dx= 0 so \int_0^x e^xf(x)dx= 2(e- 1).

That is, the integral is 2(e^x-1) for 0\le x\le 1 and the constant 2(e- 1) for x\ge 1.

 

[mehtamonica]

Thank a lot for this guidance, but i still have a doubt and hope that you may help me out this time as well. when f(x)=0 for x>=1 then how do we substitute x=1 in the solution obtained for 0<=x<1? please enlighten me on this