# Evaluate the laplace transform for this (a)L\{4x^2-3\cos(2x)+5e^x\} (b)L\{4+\frac{1}{2}\sin(x)-e^{4x}\}

Evaluate the laplace transform for this
a)$L\left\{4{x}^{2}-3\mathrm{cos}\left(2x\right)+5{e}^{x}\right\}$
b)$L\left\{4+\frac{1}{2}\mathrm{sin}\left(x\right)-{e}^{4x}\right\}$

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Step 1 Introduction
Use following laplace formulas in solving following problems
1) $L\left\{1\right\}=\frac{1}{s}$
2) $L\left\{{x}^{n}\right\}=\frac{n!}{{s}^{n+1}}$
3) $L\left\{\mathrm{cos}ax\right\}=\frac{s}{{s}^{2}+{a}^{2}}$
4) $L\left\{\mathrm{sin}ax\right\}=\frac{a}{{s}^{2}+{a}^{2}}$
5) $L\left\{{e}^{ax}\right\}=\frac{1}{s-a}$
Step 2 Step-by-Step Explanation
a)$L\left\{4{x}^{2}-3\mathrm{cos}\left(2x\right)+5{e}^{x}\right\}$
by the linearity property
$=4L\left\{{x}^{2}\right\}-3L\left\{\mathrm{cos}\left(2x\right)\right\}+5L\left\{{e}^{x}\right\}$
$=4\left(\frac{2!}{{s}^{3}}\right)-3\left(\frac{s}{{s}^{2}+{2}^{2}}\right)+5\left(\frac{1}{s-1}\right)$
$=\frac{8}{{s}^{3}}-\frac{3s}{{s}^{2}+4}+\frac{5}{s-1}$
b) $L\left\{4+\frac{1}{2}\mathrm{sin}\left(x\right)-{e}^{4x}\right\}$
by the linearity property
$=4L\left\{1\right\}+\frac{1}{2}L\left\{\mathrm{sin}\left(x\right)\right\}-L\left\{{e}^{4x}\right\}$
$=4\left(\frac{1}{s}\right)+\frac{1}{2}\left(\frac{1}{{s}^{2}+1}\right)-\frac{1}{s-4}$
$=\frac{4}{s}+\frac{1}{2\left({s}^{2}+1\right)}-\frac{1}{s-4}$
Result:
a) $L\left\{4{x}^{2}-3\mathrm{cos}\left(2x\right)+5{e}^{x}\right\}=\frac{8}{{s}^{3}}-\frac{3s}{{s}^{2}+4}+\frac{5}{s-1}$
b) $L\left\{4+\frac{1}{2}\mathrm{sin}\left(x\right)-{e}^{4x}\right\}=\frac{4}{s}+\frac{1}{2\left({s}^{2}+1\right)}-\frac{1}{s-4}$