Question

Continuing the method used in derivingand, verify the Laplace transforms of higher-order derivatives ofgiven in the table (L35).

Short answer

The Laplace transforms of higher-order derivatives of y given in the table is

$L\left(y\text{'}\right)=pY-y_0$.

Step-by-step solution

Given information

The given information is y (higher order derivatives)

Definition of Laplace Transformation

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)

Differentiate the given function

By the definition of Laplace transform

$$\begin{gathered} L(y\text{'})={\int }_0^{e^{-pt}}\frac{dy}{dx}dt \\ ={\overline{)uv}}_0-{\int }_{0}vdt \\ =e^{-pt}y{\left(t\right)}_{d-0}^{+}\left[-pt-pt\right]dt \\ ={\underset{t\to \infty 0}{lime}}^{-p{y}^p}y\left(t\right)-e^{-p0}y\left(0\right)-{\int }_0^p\left[-p{t}^{-pt}\right]y\left(t\right)dt \\ L(y\text{'})=\underset{d^{dx\to 0}}{lim}{e}^{-pt}y\left(t\right)-y\left(0\right)+p{\int }_{0}y\left(t\right)e^{-pt}dt \end{gathered}$$

If y is of exponential order $p_0$, then $\underset{l\mathit{\to }\mathit{+}}{{\mathrm{lime}}^{-\mathrm{pt}}}y\left(f\right)=0$Whenever p > p0, then

$$\begin{gathered} L(y\text{'})=0-y\left(0\right)+p{\int }_{0}y\left(t\right)e^{-pl}dt \\ =-y\left(0\right)+pL\left\{y\right(t\left)\right\}n \\ =pY-y0L(y\text{'}) \\ =pY-y0.....\left(1\right) \end{gathered}$$

Calculate the value of L (y')

$$\begin{gathered} L(y\mathit{"})=L\left(\left(y\mathit{\text{'}}\right)\text{'}\right) \\ =pL(y\text{'})-y_o^{\text{'}} \end{gathered}$$

Substitute the value in equation (1)

$$\begin{gathered} L(y\mathit{"})=p\left[pY-y_0\right]-y_0^{\text{'}} \\ =p^{2}Y-p{y}_0-y_0^{\text{'}} \\ L(y\mathit{"})=p^{2}Y-p{y}_0-y_0^{\text{'}} \end{gathered}$$ …… (2)

Calculate the value of L (y"').

$$\begin{gathered} L(y\mathit{"})=L\left(\right(y")\text{'}) \\ =pL(y")-y_0^{"} \end{gathered}$$

Substitute the value in equation (1)

$L\left(y^{\left(n\right)}\left(t\right)\right)=p^{n}Y-p^{n}Y-p^{\epsilon -1}{y}_0-p^{\epsilon -2}{y}_0^g-p^{n-3}{y}_0^n..........p{y}^{\left(n-2\right)}\left(0\right)-y^{\left(c-1\right)}\left(0\right)$

Proof for Mathematical Induction

Prove for the mathematical induction is given by

$$\begin{gathered} L\left(y^{\left(n\right)}\left(t\right)\right)=L\left({\left[y^{n-1}\left(t\right)\right]}^{\text{'}}\right) \\ =pL\left[y^{\left(\pi -1\right)}\left(t\right)\right]-y_0 \\ =p\left(p^{n-1}Y-\sum _{j=1}^{n-1}{p}^{j-1}{y}_0^{\left(\pi -1-j\right)}\right)-y_0 \\ =p^{n}Y-\sum _{j=1}^{n-1}{p}^{j-1+1}{y}_0^{\left(n-1\pi -\lambda \right)}-p^0{y}_{0}L\left(y^{\left(*\right)}\left(t\right)\right) \end{gathered}$$$$\begin{gathered} L\left(y^{\left(n\right)}\left(t\right)\right)=L\left({\left[y^{\left(n-1\right)}\left(t\right)\right]}^{\text{'}}\right) \\ =pL\left[y^{\left(\pi -1\right)}\left(t\right)\right]-y_{} \\ =p\left(p^{n-1}Y-\sum _{j=1}^{n-1}{p}^{j-1}{y}_0^{\left(\pi -1-j\right)}\right)-y_0 \\ =p^{n}Y-\sum _{j=1}^{n-1}{p}^{j-1+1}{y}_0^{\left(n-1-\lambda \right)}-p^0{y}_{0}L\left(y^{\left(*\right)}\left(t\right)\right) \end{gathered}$$

$$\begin{gathered} L\left(y^{\left(n\right)}\left(t\right)\right)=p^{n}Y-\sum _{j=1}^{\pi -1}{p}^i{y}_0^{\left(n-j\right)} \\ =p^{n}Y-\sum _{j=1}^n{p}^{-1}{y}_0^{\left(n-\lambda \right)} \end{gathered}$$