- #1
paraboloid
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Solve the following given y(0) = 0 & y'(0)=1:
y′′+3y′+2y = u2(t), such that u2(t) is a heaviside step function
Here's what I've got so far,
=>s2Y(s)−sy(0)−y′(0) + 3sY(s)−3y(0) + 2Y(s)= exp(−2s)/s
Y(s) = (exp(−2s) + s) / (s(s2+3s+2))
Y(s) = exp(−2s)/(s(s2+3s+2))* + 1/(s2+3s+2)**
The second part, **, I was able to solve with partial fractions => 1/(s+1) − 1/(s+2) which transforms to exp(−t) − exp(−2t).
However I don't know how to solve the first part, *, since the step function isn't by itself,
Any push in the right direction would be great,
Thanks in advance
y′′+3y′+2y = u2(t), such that u2(t) is a heaviside step function
Here's what I've got so far,
=>s2Y(s)−sy(0)−y′(0) + 3sY(s)−3y(0) + 2Y(s)= exp(−2s)/s
Y(s) = (exp(−2s) + s) / (s(s2+3s+2))
Y(s) = exp(−2s)/(s(s2+3s+2))* + 1/(s2+3s+2)**
The second part, **, I was able to solve with partial fractions => 1/(s+1) − 1/(s+2) which transforms to exp(−t) − exp(−2t).
However I don't know how to solve the first part, *, since the step function isn't by itself,
Any push in the right direction would be great,
Thanks in advance