Question

By using Laplace transforms, solve the following differential equations subject to the given initial conditions.

$\mathit{y}\mathbf{"}\mathbf{}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{4}{\mathit{e}}^{\mathbf{2}\mathbf{t}}\mathbf{,}\mathbf{}\mathbf{\sum }_{\mathbf{}}{\mathit{y}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}{\mathbf{\text{'}}}_{\mathbf{0}}\mathbf{=}\mathbf{1}$

Short answer

The solution of given differential equation is y =$t{e}^{2t}$.

Step-by-step solution

Given information

The equation is$y"-4y=4e^{2t}\mathrm{and}{\mathrm{y}}_0=0,\mathrm{y}{\text{'}}_0=1$.

 Step 2: 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.

$y"-4y=4e^{2t}$

Take the Laplace of above equation.

$$\begin{gathered} L\left\{y"-4y\right\}=L\left\{4e^{2i}\right\}L\left\{y"\right\}-4L\left\{y\right\} \\ =4L\left\{e^{2t}\right\}{p}^{2}L\left\{y\right\}-p{y}_0-y{\text{'}}_0-4L\left\{y\right\} \\ =\frac{4}{p-2}\left(p^2-4\right)L\left\{y\right\}-p{y}_0-y{\text{'}}_0 \\ =\frac{4}{p-2} \end{gathered}$$

Further solve,

$$\begin{gathered} \left(p^2-4\right)L\left\{y\right\}-p\left(0\right)-1=\frac{4}{p-2}\left(p^2-4\right)L\left\{y\right\} \\ =\frac{4}{p-2}+1L\left\{y\right\} \\ =\frac{4}{p-2\left(p^2-4\right)}+\frac{1}{\left(p^2-4\right)}L\left\{y\right\} \end{gathered}$$

The inverse Laplace is,

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

Thus, the solution of given differential equation is $y=t{e}^{2t}$.