Solving ODE with Heaviside Step and Delta function

  • #1

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 dont know how to continue

please help guys...thanks
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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Do you know what those functions are? The 'Heaviside step function' is 0 if x< 0, 1 if [itex]x\ge 0[/itex]. So [itex]\int H(x)e^{-t/a} dx= 0[/itex] if x is negative and [itex]\int e^{-t/a} dt[/itex] 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), [itex]\int f(x)\delta(x)dx= f(0)[/itex].
 

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