Question

In problem and use the Laplace transform and these inverse to solve the given initial-value problem.

${\mathit{y}}^{\mathbf{n}}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{,}\mathit{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{3}$

Short answer

The solved initial-value problem use the Laplace transform is$L\left[y\right]=e^{t}cos\left(2t\right)+e^{t}sin\left(2t\right)$

Step-by-step solution

Step1: definition of Laplace transformation.

A transformation of a function f(x) into the function is particularly useful for reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

$$\begin{gathered} s^{2}L\left[y\right]-sy\left(0\right)-y\text{'}\left(0\right)-2\left(sL\left[y\right]-y\left(0\right)\right)+5L\left[y\right]=0 \\ {s}^{2}L\left[y\right]-s-3-2\left(sL\left[y\right]-1\right)+5L\left[y\right]=0 \\ \begin{array}{c}L\left[y\right]\left(s^2-2s+5\right)=s+1\\ L\left[y\right]=\frac{s+1}{s^2-2s+5}\end{array} \end{gathered}$$

In the preceding procedure, we used the following formula:

$L\left[f^{\left(n\right)}\left(t\right)\right]=s^{n}L\left[f\left(t\right)\right]-s^{n-1}f\left(0\right)-s^{n-2}f\text{'}\left(0\right)-..-f^{\left(n-1\right)}\left(0\right)$

define linearity and formulas:

Linearity refers to the property of a mathematical relationship (function) that can be represented graphically as a straight line. Linearity and proportionality are inextricably linked.

$\begin{array}{l}L\left[e^{at}cos\left(bt\right)\right]=\frac{s-a}{{\left(s-a\right)}^2+b^2}\\ L\left[e^{at}sin\left(bt\right)\right]=\frac{b}{{\left(s-a\right)}^2+b^2}\end{array}$

Then we have,

$L\left[y\right]=\frac{s+1}{s^2-2s+5}=\frac{s+1}{{\left(s-1\right)}^2+4}=\frac{s-1}{{\left(s-1\right)}^2+2^2}+\frac{2}{{\left(s-1\right)}^2+2^2}$

$\begin{array}{c}y\left(t\right)=L^{-1}\left[\frac{s-1}{{\left(s-1\right)}^2+2^2}+\frac{2}{{\left(s-1\right)}^2+2^2}\right]\\ =L^{-1}\frac{s-1}{{\left(s-1\right)}^2+2^2}+L^{-1}\frac{2}{{\left(s-1\right)}^2+2^2}\end{array}$

$e^t\cos \left(2t\right)+e^t\sin \left(2t\right)$

Hence,

$L\left[y\right]=e^{t}cos\left(2t\right)+e^{t}sin\left(2t\right)$