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Laplace transform of e^{at}

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rebelmaths

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Note that the Laplace transform of $e^{at}$ is $\\frac{1}{s-a}$

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$L[e^{at}]=\\frac{1}{s-a}$

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Use this result to solve the following:

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Recall that $L[e^{at}]=\\frac{1}{s-a}$

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(a)

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Using the above result, $L[e^{\\var{a}t}]=\\frac{1}{s-\\var{a}}$

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(b)  

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Using the above result,  $L[e^{\\var{b}t}]=\\frac{1}{s-(\\var{b})}=\\frac{1}{s+\\var{-b}}$

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(c)

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When you have the Laplace transform of two functions added together you just get the Laplace transform of each function and add the two answers.

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$L[f(t)+g(t)]=L[f(t)]+L[g(t)]$

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Hence  $L[e^{\\var{c}t}+e^{\\var{d}t}]=L[e^{\\var{c}t}]+L[e^{\\var{d}t}]=\\frac{1}{s-(\\var{c})}+\\frac{1}{s-\\var{d}}=\\simplify{1/(s-{c})+1/(s-{d})}$

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Find the Laplace transform of $e^{\\var{a}t}$

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Find the Laplace transform of $e^{\\var{b}t}$

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When you have the Laplace transform of two functions added together you just get the Laplace transform of each function and add the two answers.

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$L[f(t)+g(t)]=L[f(t)]+L[g(t)]$

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Find the Laplace transform of $ { e^{ \\var{c} t}+e^{ \\var{d} t} }$

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