Solving ODE with Heaviside Step and Delta function

[dominic.tsy]

Homework Statement

Find the solution of the equation:

α(dy/dt) + y = f(t)

for the following conditions:
(a) when f(t) = H(t) where H(t) is the Heaviside step function
(b) when f(t) = δ(t) where δ(t) is the delta function
(c) when f(t) = β^(-1)e^(t/β)H(t) with β<α

Homework Equations

The Attempt at a Solution

1.
α(dy/dt) + y = f(t)
2.
(dy/dt) + (1/α)y = (1/α)f(t)
3.
finding the integrating factor
μ(t) = e^(∫(1/α)dt) = e^(t/α)
4.
[e^(t/α)](dy/dt) + (1/α)[e^(t/α)]y = (1/α)[e^(t/α)]f(t)
5.
d/dt{[e^(t/α)]y}=[(e^(t/α))/α]f(t)
6.
∫d/dt{[e^(t/α)]y}dt=∫[(e^(t/α))/α]f(t)dt
7.
[e^(t/α)]y = ...

then i don't know how to continue

please help guys...thanks

 

[HallsofIvy]

Do you know what those functions are? The 'Heaviside step function' is 0 if x< 0, 1 if x\ge 0. So \int H(x)e^{-t/a} dx= 0 if x is negative and \int e^{-t/a} dt if x is positive.

The 'delta function' (strictly speaking not a function but a 'generalized function' or 'distribution') has the property that for any function, f(x), \int f(x)\delta(x)dx= f(0).