Question

In problem 21-30use the Laplace transform to solve the givenInitial-value Problem.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{=}\mathbf{1}\mathbf{+}\mathit{t}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{4}$

Short answer

The Laplace transform to solve the given

Initial-value Problem.

$y\text{'}\text{'}-2y\text{'}+5y=1+t,y\left(0\right)=0,y\text{'}\left(0\right)=4$

$\begin{array}{c}f\left(t\right)=e^t-e^{-t^2}\\ =e^{2t}-2+e^{-2t}\end{array}$

$L\left[f\left(t\right)\right]==\frac{1}{S-2}-\frac{2}{S}+\frac{1}{S+2}$

Step-by-step solution

definition of Laplace transforms

A transformation of a function is particularly useful in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

$y\text{'}\text{'}-2y\text{'}+5y=1+t,y\left(0\right)=0,y\text{'}\left(0\right)=4$

Take both sides' Laplace transforms.

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

Step 2: procedure Used the linearity

Linearity is the property of a mathematical relationship that can be represented graphically as a straight line..

In the preceding procedure,

we used the linearity of the Laplace transform and the following formulas:

$$\begin{gathered} \begin{array}{l}L\left[1\right]=\frac{1}{S}\\ L{e}^{at}=\frac{1}{S-a}\end{array} \\ L\left[f\left(t\right)\right]==\frac{1}{S-2}-\frac{2}{S}+\frac{1}{S+2} \end{gathered}$$

Hence,

$L\left[f\left(t\right)\right]==\frac{1}{S-2}-\frac{2}{S}+\frac{1}{S+2}$