How Do You Analyze a Continuous Time LTI System Using Differential Equations?

[crom1]

Homework Statement

continuous time LTI system is given with differential equation y'(t) + 5y(t) = 10u(t)
a) Find transfer function of system and determine is it stable or not.
b)Find frequency characteristics (amplitude and phase angle) of given system
c) Find impulse response using Laplace transformation
d) Find zero-input response(I think that's the right word) for u(t) = 6cos(5t)μ(t)

The Attempt at a Solution

Ok, I got
a) Y(s) = 10/(s+5) * U(s) , pole is s=-5 , and Re(s)<0 so the system is stable
b) Plug s=jω in H(s)=10/(s+5) , then |H(jω)| = 10/(√s^2+ω^2) and ∠H(jω) = - arctan(ω/5)
c) for impulse response U(s) = 1, Y(s) = 10/(s+5) ⇒ y(t) = 10*e^(-5t) , t>0
d) Using frequency characteristics, we have ω=5, and |H(j5)| = √2 and ∠H(j5) = -π/4 so the response is
y(t) = 6√2 *cos(5t-pi/4)

I need someone to say if these are correct or not, if not i will post my full attempt and reasoning. Thanks

 

[rude man]

Ok, I got
a) Y(s) = 10/(s+5) * U(s) , pole is s=-5 , and Re(s)<0 so the system is stable

10 U(s) is not part of the transfer function. It's a specific input; the transfer function gives output for any input.

b) Plug s=jω in H(s)=10/(s+5) , then |H(jω)| = 10/(√s^2+ω^2) and ∠H(jω) = - arctan(ω/5)

OK but leave out the "10". Also, typo used "s" for "5".

c) for impulse response U(s) = 1, Y(s) = 10/(s+5) ⇒ y(t) = 10*e^(-5t) , t>0

U is a step function.The symbol for impulse input is δ(t). Your actual input is kδ(t), k = 1 volt-sec. Careful with δ(t), its units are time^-1. Again, ditch the "10".

d) Using frequency characteristics, we have ω=5, and |H(j5)| = √2 and ∠H(j5) = -π/4 so the response is
y(t) = 6√2 *cos(5t-pi/4)

never heard of "zero input response". You mean mean input is 6cos(5t)U(t) or ? Never saw μ(t) either.
Strange they never asked for the time response to the 10u(t) input ...