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

${\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{\text{'}}\mathbf{-}\mathbf{8}{\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{16}\mathbf{y}\mathbf{=}\mathbf{32}\mathbf{t}\mathbf{,}\mathbf{\hspace{1em}}{\mathbf{y}}_{\mathbf{0}}\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{y}}_{\mathbf{0}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{2}$

The differential equation is${\mathrm{y}}^{\text{'}}\text{'}-8{\mathrm{y}}^{\text{'}}+16\mathrm{y}=32\mathrm{t},\hspace{1em}{\mathrm{y}}_0=1,{\mathrm{y}}_0^{\text{'}}=2.$ .

Application of Laplace transform:

It's used to simplify complicated differential equations by using polynomials. It's used to convert derivatives into numerous domain variables, then utilise the Inverse Laplace transform to convert the polynomials back to the differential equation.

The given equation is

${\mathrm{y}}^{\text{'}}\text{'}-8{\mathrm{y}}^{\text{'}}+16\mathrm{y}=32\mathrm{t}$

Take the Laplace of equation

localid="1659249826543" $$\begin{gathered} \mathrm{L}\left\{{\mathrm{y}}^{\text{'}}\text{'}-8{\mathrm{y}}^{\text{'}}+16\mathrm{y}\right\}=\mathrm{L}\left\{32\mathrm{t}\right\} \\ \mathrm{L}\left\{\mathrm{y}"\right\}-8\mathrm{L}\left(\mathrm{y}\text{'}\right)+16\mathrm{L}\left\{\mathrm{y}\right\}=32\mathrm{L}\left\{\mathrm{t}\right\} \\ {\mathrm{p}}^2\mathrm{L}\left\{\mathrm{y}\right\}-{\mathrm{py}}_0-{\mathrm{y}}_0\text{'}-8\left[\mathrm{pL}\left\{\mathrm{y}\right\}-{\mathrm{y}}_0\right]+16\mathrm{L}\left\{\mathrm{y}\right\}=\frac{32}{{\mathrm{p}}^2} \\ \left({\mathrm{p}}^2-8\mathrm{p}+16\right)\mathrm{L}\left\{\mathrm{y}\right\}+\left(8-\mathrm{p}\right){\mathrm{y}}_0-\mathrm{y}{\text{'}}_0=\frac{32}{{\mathrm{p}}^2} \end{gathered}$$

Further solve

$$\begin{gathered} \left({\mathrm{p}}^2-8\mathrm{p}+16\right)\mathrm{L}\left\{\mathrm{y}\right\}+\left(8-\mathrm{p}\right)\left(1\right)-2=\frac{32}{{\mathrm{p}}^2} \\ \left({\mathrm{p}}^2-8\mathrm{p}+16\right)\mathrm{L}\left\{\mathrm{y}\right\}=\frac{32}{{\mathrm{p}}^2}-6+\mathrm{p} \\ \mathrm{L}\left\{\mathrm{y}\right\}=\frac{32}{{\mathrm{p}}^2\left({\mathrm{p}}^2-8\mathrm{p}+16\right)}-\frac{6}{\left({\mathrm{p}}^2-8\mathrm{p}+16\right)}+\frac{\mathrm{p}}{\left({\mathrm{p}}^2-8\mathrm{p}+16\right)} \\ \mathrm{L}\left\{\mathrm{y}\right\}=\frac{1}{\mathrm{p}}+\frac{2}{{\mathrm{p}}^2} \end{gathered}$$

The inverse Laplace is

$$$\begin{gathered} y=L^{-1}\frac{1}{p}+L^{-1}\left\{\frac{2}{p^2}\right\} \\ y=L^{-1}\left\{\frac{1}{p}\right\}+2L^{-1}\left\{\frac{1}{p^2}\right\} \\ y=1+2t \end{gathered}$$

Thus, the given differential equation's solution is$y=1+2t$.