thundercleese
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I get close to the answer but just not seeing where the integers in the answers are coming from, perhaps forgetting a constant? This is a EE circuit analysis class.
Given y(t) = dx(t)/dt
a. Find x(t) if y(t)=
Part a
y(t)=\frac{dx(t)}{dt}
when y(t) = 0:
∫0 = ∫\frac{dx(t)}{dt}
0 = x(t) (this part agrees with given answer)
when y(t) = e^{-t}:
∫ e^{-t}=∫\frac{dx(t)}{dt}
-e^{-t}=x(t)
part b, y(t)=t^{2}:
∫t^{2} = ∫\frac{dx(t)}{dt}
\frac{1}{3}t^{3} = x(t)
Given answers:
part a: 0, t\leq0; (1-e^{-t}), t\geq0
part b: \frac{1}{3}(2+t^{3}), t \geq1
Homework Statement
Given y(t) = dx(t)/dt
a. Find x(t) if y(t)=
0 : t<0;
e^{-t} : t>0
b. Find x(t) for t=1 if x(1)=1 and y(t)=t^{2} for t\geq1The Attempt at a Solution
Part a
y(t)=\frac{dx(t)}{dt}
when y(t) = 0:
∫0 = ∫\frac{dx(t)}{dt}
0 = x(t) (this part agrees with given answer)
when y(t) = e^{-t}:
∫ e^{-t}=∫\frac{dx(t)}{dt}
-e^{-t}=x(t)
part b, y(t)=t^{2}:
∫t^{2} = ∫\frac{dx(t)}{dt}
\frac{1}{3}t^{3} = x(t)
Given answers:
part a: 0, t\leq0; (1-e^{-t}), t\geq0
part b: \frac{1}{3}(2+t^{3}), t \geq1