Use L29 to verify L6, L13, L14, and L18

The given function is$L29$

A transformation of a function f(x) into the function g(t) 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(s) is the piecewise-continuous and exponentially-restricted real function f(t)

From (1.29).

$L\left\{e^{-a{R}_g\left(t\right)}\right\}=G\left(p+a\right)$

From (L5)

$L\left(t^k\right)=\frac{1}{p^{d+1}}$ …… (1)

Consider $g\left(t\right)=t^k$and use $\left(L29\right),L\left\{e^{-a{t}_g}\left(t\right)\right\}=G\left(p+a\right);L\left\{t^k{e}^{-a}\right\}$can be calculated by replace $p$by $\left(p+a\right)$in equation (1) as,

$$\begin{gathered} L\left(t\right)=\frac{k!}{p^{i+1}} \\ L\left\{t^k{e}^{-at}\right\}=\frac{k!}{{\left\{p+a\right\}}^{k+1}} \end{gathered}$$

Hence, $\left(L6\right):L\left\{t^k{e}^{-w}\right\}=\frac{k!}{{\left(p+w\right)}^{t+1}}$is verified.

From $\left(.29\right)$.

$L\left\{e^{-a{t}_g\left(f\right)}\right\}=G\left(p+a\right)$

From $\left(l,3\right)$.

$L\left\{\sin bt\right\}=\frac{b}{p^2+b^2}$ …… (2)

Consider $g\left(t\right)=\sin bt$$\left(L29\right),L\left\{e^{{-}^{-w}\left(t\right)}\right\}=G\left(p+a\right);L\left\{e^{-at}\sin bt\right\}$and use can be calculated by replace p by$\left(p+a\right)$ in equation (2) .

$L\left\{\sin bt\right\}=\frac{b}{p^2+b^2}L\left\{e^{-d}\sin bt\right\}=\frac{b}{{\left(p+a\right)}^2+b^2}$Hence,$\left(L13\right)$$L\left\{e^{-w}\sin br\right\}=\frac{b}{{\left(p+a\right)}^2+b^2}$is verified

From$\left(L29\right)$.

$L\left\{e^{-a{t}_g\left(f\right)}\right\}=G\left(p+a\right)$

From$\left(L3\right)$.

$L\left\{\cos bt\right\}=\frac{b}{p^2+b^2}$ …… (3)

Consider $g\left(t\right)=\sin bt$$\left(L29\right),L\left\{e^{-a}g\left(t\right)\right\}=G\left(p+a\right);L\left\{e^{-at}\cos bt\right\}$and use can be calculated by replace p by $\left(p+a\right)$in equation.

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

Hence,role="math" localid="1659341085564" $\left(L14\right);$role="math" localid="1659341000985" $L\left\{e^{-w^t}\cos bt\right\}=\frac{b}{{\left(p+a\right)}^2+b^2}$is verified

From property (LI) of Laplace transform,

$L\left(t\right)=\frac{1}{F}$ …… (4)

From property (L.S) of Laplace transform,

$$\begin{gathered} L\left(t^k\right)=\frac{k}{p^{i+1}} \\ L\left(t\right)=\frac{1}{p^2} \\ L\left(at\right)=\frac{a}{p^2} \end{gathered}$$ …… (5)

Subtract equation from equation (5) from (4) as,

$$\begin{gathered} L\left(r\right)-L\left(a\right)=\frac{1}{p}-\frac{a}{p^2} \\ L\left(1-af\right)=\frac{p-a}{p^2} \end{gathered}$$

From$\left(L29\right)$.

$L\left\{e^{-a{t}_g\left(t\right)}\right\}=G\left(p+a\right)$

Since,

$L\left(1-at\right)=\frac{p-a}{p^2}$

Consider $g\left(t\right)=\left(1-at\right)$and use $\left(L29\right).L\left\{e^{-w}\left(1-t\right)\right\}$can be calculated by replace p by $\left(p+a\right)$in equation (3).

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

Hence, $\left(L\right)8);L\left\{e^{-w}\left(1-t\right)\right\}=\frac{p}{{\left(p+a\right)}^2}$is verified.

Thus, the values of L6, L13, LI4 and L 18 using is and

$$\begin{gathered} \left(L.6\right):L\left\{t^k{e}^{-at}\right\}=\frac{k}{{\left(p+a\right)}^{k+1}} \\ \left(L13\right):L\left\{e^{-at}\sin bt\right\}=\frac{b}{{\left(p+a\right)}^2+b^2}, \\ \left(L14\right):L\left\{e^{-at}\cos br\right\}=\frac{b}{{\left(p+a\right)}^2+b^2}, \\ \left(L18\right):L\left\{e^{-at}\left(1-t\right)\right\}=\frac{p}{{\left(p+a\right)}^2}. \end{gathered}$$