- #1
dominic.tsy
- 6
- 0
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
My try for all 3 are as follow:
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
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
My try for all 3 are as follow:
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