Asked by dam
show that
2f''(t)-f'(t)-f(t)=sin(t)-cos(t)
where
f(0)=f'(0)
has the solution
(-2/5)e^(-t/2)-(1/5)sin(t)+(2/5)cos(t)
using laplace tranformation
plz plz plz show step by step
2f''(t)-f'(t)-f(t)=sin(t)-cos(t)
where
f(0)=f'(0)
has the solution
(-2/5)e^(-t/2)-(1/5)sin(t)+(2/5)cos(t)
using laplace tranformation
plz plz plz show step by step
Answers
Answered by
Steve
L{f'(t)} = sF(s)-f(0)
L{f"(t)} = s^2 F(s) - sf(0) - f'(0)
Not sure what to do with f(0)=f'(0) since you don't give a value.
L{sin(t)-cos(t)} = (1-s)/(1+s^2)
Just plug in the expressions and it will all fall out.
L{f"(t)} = s^2 F(s) - sf(0) - f'(0)
Not sure what to do with f(0)=f'(0) since you don't give a value.
L{sin(t)-cos(t)} = (1-s)/(1+s^2)
Just plug in the expressions and it will all fall out.
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