document.write( "Question 1185331: Find the inverse Laplace transform f(t)=L^(−1){F(s)} of the function
\n" ); document.write( "F(s)=(e^(−s)(6s−5))/(s^2+64)
\n" ); document.write( "You may use h(t) for the Heaviside step function.\r
\n" ); document.write( "\n" ); document.write( "f(t)=L^(−1){(e^(−s)(6s−5))/(s^2+64)}=\r
\n" ); document.write( "\n" ); document.write( "My answer is h(t-1)(6cos(8h(t-1)))-5/8sin(8h(t-1)) , wrong \n" ); document.write( "
Algebra.Com's Answer #849761 by CPhill(1959)\"\" \"About 
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Here's how to find the inverse Laplace transform of F(s) = (e^(-s)(6s - 5))/(s² + 64):\r
\n" ); document.write( "\n" ); document.write( "1. **Recognize the time shift:** The term e^(-s) indicates a time shift. Recall that L{f(t-a)u(t-a)} = e^(-as)F(s), where u(t-a) is the Heaviside step function. In our case, a = 1. This means we'll have a function shifted by 1 unit in time.\r
\n" ); document.write( "\n" ); document.write( "2. **Inverse transform of the rest:** We need to find the inverse Laplace transform of G(s) = (6s - 5)/(s² + 64). We can break this into two parts:\r
\n" ); document.write( "\n" ); document.write( " * L⁻¹{6s/(s² + 64)} = 6L⁻¹{s/(s² + 8²)} = 6cos(8t)
\n" ); document.write( " * L⁻¹{-5/(s² + 64)} = -5L⁻¹{1/(s² + 8²)} = (-5/8)L⁻¹{8/(s² + 8²)} = (-5/8)sin(8t)\r
\n" ); document.write( "\n" ); document.write( " Therefore, L⁻¹{G(s)} = 6cos(8t) - (5/8)sin(8t)\r
\n" ); document.write( "\n" ); document.write( "3. **Apply the time shift:** Since F(s) = e^(-s)G(s), we have f(t) = g(t-1)u(t-1), where g(t) = 6cos(8t) - (5/8)sin(8t).\r
\n" ); document.write( "\n" ); document.write( " So, f(t) = [6cos(8(t-1)) - (5/8)sin(8(t-1))]u(t-1)\r
\n" ); document.write( "\n" ); document.write( "Therefore, the inverse Laplace transform is:\r
\n" ); document.write( "\n" ); document.write( "f(t) = [6cos(8(t-1)) - (5/8)sin(8(t-1))]u(t-1)
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