Find the Laplace transforms of the following function:
4. L{e^5t}
L(e5t)=∫0∞e−ste5tdt=∫0∞e(5−s)tdt=15−slimx→∞e(5−s)t∣0x=15−slimx→∞(e(5−s)x−e0)=15−s(0−1)=1s−5L({e^{5t}}) = \int\limits_0^\infty {{e^{ - st}}{e^{5t}}} dt = \int\limits_0^\infty {{e^{\left( {5 - s} \right)t}}} dt = \frac{1}{{5 - s}}\mathop {\lim }\limits_{x \to \infty } \left. {{e^{\left( {5 - s} \right)t}}} \right|_0^x = \frac{1}{{5 - s}}\mathop {\lim }\limits_{x \to \infty } \left( {{e^{\left( {5 - s} \right)x}} - {e^0}} \right) = \frac{1}{{5 - s}}\left( {0 - 1} \right) = \frac{1}{{s - 5}}L(e5t)=0∫∞e−ste5tdt=0∫∞e(5−s)tdt=5−s1x→∞lime(5−s)t∣∣0x=5−s1x→∞lim(e(5−s)x−e0)=5−s1(0−1)=s−51
Answer: 1s−5\frac{1}{{s - 5}}s−51
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