Question

By using Laplace transforms, solve the following differential equations subject to the given initial conditions.${\mathit{y}}^{\mathbf{t}}\mathbf{-}\mathit{y}\mathbf{=}\mathbf{2}{\mathit{t}}^{\mathbf{t}}\mathbf{}\mathbf{,}\mathbf{}\mathbf{2}\mathit{t}\mathbf{=}\mathbf{3}$

Short answer

Answer

The solution of given differential equation is$y-e^t\left(3+2t\right)$.

Step-by-step solution

Given information

The given equation is $y-y=2e^t$and$y_0-3$.

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

Consider the equation

$y^{\text{'}}-y=2e^x$

Take the Laplace of above equation.

$$\begin{gathered} L\left\{y-y\right\}=L\left\{2,\right\}n \\ L\left\{y\text{'}\right\}-L\left\{y\right\}=L\left\{2e\right\} \end{gathered}$$

Use $L\left\{e^w\right\}=\frac{1}{p+a}$and $L\left\{y\text{'}\right\}=pL\left\{y\right\}-y$in above equation.

$$\begin{gathered} pL\left\{y\right\}-y_0-L\left\{y\right\}=\frac{2}{p-1} \\ pL\left\{y\right\}-3-L\left\{y\right\}=\frac{2}{p-1} \\ L\left\{y\right\}\left(p-1\right)=\frac{3p-1}{p-1} \\ L\left\{y\right\}=\frac{3p-1}{{\left(p-1\right)}^2} \end{gathered}$$

The inverse Laplace is,

$$\begin{gathered} y=L^{-1}\left\{\frac{3p-1}{{\left(p-1\right)}^2}\right\} \\ =3L^{-1}\left\{\frac{p}{{\left(p-1\right)}^2}\right\}-L^{-1}\left\{\frac{1}{{\left(p-1\right)}^2}\right\} \\ =3e^2\left(1+t\right)-t{e}^{2}y \\ =e\text{'}\left(3+2t\right) \end{gathered}$$

Thus, the solution of given differential equation is$y=e^t\left(3+2t\right)$.