Asked by Sara
                Use the fact that L{f'}=sF-f(0) to derive the formula for L{cosh at}
            
            
        Answers
                    Answered by
            Steve
            
    cosh(at) = d/dt (1/a sinh(at))
sinh(at) = d/dt (1/a cosh(at))
So,
Since L{f(at)} = 1/|a| F(s/a)
L{cosh(at)} = sL{1/a sinh(t)}
= s/a L{sinh(t)}
L{sinh(at)} = sL{1/a cosh(at)}
= s/a L{cosh(at)} - s/a
So,
L{cosh(at)} = s/a (s/a L{cosh(at)} - s/a
L{cosh(at)}(1-s^2/a^2) = -s/a
= (s/a^2)/(s^2/a^2 - 1)
= s/(s^2-a^2)
Better check my algebra in the middle steps. I think I dropped an "a" somewhere, but the final answer is correct. You work this in the same way you do integration by parts twice for trig functions.
    
sinh(at) = d/dt (1/a cosh(at))
So,
Since L{f(at)} = 1/|a| F(s/a)
L{cosh(at)} = sL{1/a sinh(t)}
= s/a L{sinh(t)}
L{sinh(at)} = sL{1/a cosh(at)}
= s/a L{cosh(at)} - s/a
So,
L{cosh(at)} = s/a (s/a L{cosh(at)} - s/a
L{cosh(at)}(1-s^2/a^2) = -s/a
= (s/a^2)/(s^2/a^2 - 1)
= s/(s^2-a^2)
Better check my algebra in the middle steps. I think I dropped an "a" somewhere, but the final answer is correct. You work this in the same way you do integration by parts twice for trig functions.
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