Verify L15 to L18, by combining appropriate preceding formulas

The given functions is

$$\begin{gathered} \left(L15\right);L(1-\cos at)=\frac{a^2}{\rho \left({\rho }^2+a^2\right)} \\ \left(L16\right);L(at-\sin at)=\frac{u^1}{p^2\left({\rho }^2+a^2\right)} \\ \left(L17\right);L(\sin at-at\cos at)=\frac{2u^3}{{\left({\rho }^2+a^2\right)}^2}, \\ \left(L18\right);L\left(e^{-at}(1-at)\right)=\frac{p}{{(0+a)}^2} \end{gathered}$$

A transformation of a function $f\left(x\right)$ into the function$g\left(t\right)$ that is useful especially in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

The inverse Laplace transform of a function$F\left(s\right)$is the piecewise-continuous and exponentially-restricted real function$f\left(t\right)$.

From, property (L14) of Laplace transforms$L\left(1\right)=\frac{1}{p}\dots ..\left(1\right)$

From, property L (4) of Laplace transforms$L\left(\cos af\right)=\frac{p}{p^2+a^2}\dots \dots .\left(2\right)$

Subtract equation (2) from (1) as

$$\begin{gathered} L\left(1\right)-L\left(\cos at\right)=\frac{1}{p}-\frac{p}{p^2+a^2} \\ L(1-\cos at)=\frac{1}{p}-\frac{p}{p^2+a^2} \\ L(1-\cos at)=\frac{p^2+a^2-p^2}{p\left(p^2+a^2\right)} \\ L(1-\cos at)=\frac{a^2}{p\left({\rho }^2+a^2\right)} \end{gathered}$$

Hence,$L(1-\cos at)=\frac{a^2}{P\left({\rho }^2+u^2\right)}$

From, Property (15) of Laplace transforms

$$\begin{gathered} L\left(t^k\right)=\frac{w}{p+1} \\ L\left(t\right)=\frac{1}{p^2} \\ L\left(at\right)=\frac{a}{p^2}\dots \left(3\right) \end{gathered}$$

From, Property L(14) of Laplace transforms

$L\left(\cos af\right)=\frac{p}{v^2+a^2}\dots \dots ..\left(4\right)$

Subtract equation (4) from (3) as,

$$\begin{gathered} L\left(at\right)-L\left(\sin at\right)=\frac{a}{p^2}-\frac{a}{p^2+a^2} \\ L(at-\sin at)=\frac{u\left(p^2+a^2\right)-a{p}^2}{p^2\left(p^2+a^2\right)} \\ L(at-\sin at)=\frac{a{p}^2+a^3-w^2}{p^2\left(p^2+a^2\right)} \\ L(at-\sin at)=\frac{a^3}{p^2\left(p^2+a^2\right)} \end{gathered}$$

Hence,$L(at-\sin at)=\frac{w^3}{p^2\left(p^2+a^2\right)}$

From, Property (12) of Laplace transforms

$$\begin{gathered} L\left(t\cos at\right)=\frac{p^2-a^2}{{\left(p^2+a^2\right)}^2} \\ L\left(at\cos at\right)=\frac{e\left(p^2-a^2\right)}{{\left(p^2+a^2\right)}^2} \\ L\left(at\cos at\right)=\frac{a{p}^2-a^3}{{\left(p^2+a^2\right)}^2} \end{gathered}$$

From, Property role="math" localid="1664280923563" $L\left(3\right)$of Laplace transforms,

$L\left(a\sin at\right)=\frac{\omega }{p^2+a^2}\dots \dots \left(6\right)$

Subtract equation (5) from (6) as

$$\begin{gathered} L\left(\sin at\right)-L\left(at\cos at\right)=\frac{a}{p^2+a^2}-\frac{\left(a{p}^2-a^3\right)}{{\left(p^2+a^2\right)}^2} \\ L(\sin at-at\cos at)=\frac{e\left({\rho }^2+a^2\right)-\left({\alpha }^2-a^5\right)}{{\left(p^2+a^2\right)}^2} \\ L(\sin at-at\cos at)=\frac{4v^2+a^3-a{p}^2+a^3}{{\left(p^2+a^2\right)}^2} \\ L(\sin at-at\cos at)=\frac{2{\omega }^3}{{\left({\rho }^2+a^2\right)}^2} \end{gathered}$$

Hence,$L(\sin at-at\cos at)=\frac{2{\mu }^3}{{\left({\rho }^2+a^2\right)}^2}$

From, Property L12 of Laplace transforms is

$L\left(e^{-at}\right)=\frac{1}{ptw}\dots \dots \text{(7) A}$

From, Property L16 of Laplace transforms is

$$\begin{gathered} L\left(t^k{e}^{-\alpha t}\right)=\frac{1}{{(p+a)}^{k+1}}L\left(t{e}^{-wt}\right) \\ =\frac{1}{{(p+u)}^2}L\left({\text{ate}}^{-w}\right) \\ =\frac{a}{{(p+\omega )}^2}\dots \dots \left(8\right) \end{gathered}$$

Subtract equation 8 from 7

$$\begin{gathered} L\left(e^{-at}\right)-L\left(at{e}^{-at}\right)=\frac{1}{p+a}-\frac{w}{{(p+a)}^2} \\ L\left(e^{-at}(1-at)\right)=\frac{p^{ta-a}-a}{{(p+a)}^2} \\ L\left(e^{-at}(1-at)\right)=\frac{p}{{(p+a)}^2} \\ \text{Hence,}L\left(e^{-w}(1-at)\right)=\frac{F}{{(r+a)}^2}\text{.} \end{gathered}$$