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

${\mathit{y}}^{\mathbf{"}}\mathbf{-}\mathbf{6}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathit{y}\mathbf{=}\mathit{t}{\mathit{e}}^{\mathbf{3}\mathbf{t}}$,

The given equation is $y^{"}-6y+9y=t{e}^{3r}$and$y_0-0,y_0-5$.

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)

Consider the equation.

$y"-6y+9y=t{e}^{3t}$

Take the Laplace of above equation.

$$\begin{gathered} L\left\{y"-6y\text{'}+9y\right\}=L\left\{t{e}^{3t}\right\}L\left\{y"\right\}-6L\left\{y\text{'}\right\}+9 \\ L\left\{y\right\}=L\left\{t{e}^{3t}\right\}{p}^{2}L\left(y\right)-p{y}_0-6\left[pL\left\{y\right\}-y_0\right]+9 \\ L\left\{y\right\}=\frac{1}{{\left(p-3\right)}^2}\left(p^2-6p+9\right)L\left\{y\right\}+\left(6-p\right)y_0-y_0 \\ =\frac{1}{{\left(p-3\right)}^2} \end{gathered}$$

Further solve,

$$\begin{gathered} \left(p^2-6p+9\right)L\left\{y\right\}+\left(6-p\right)\left(0\right)-5=\left(p^2-6p+9\right) \\ L\left(y\right)=\frac{1}{{\left(p-3\right)}^2}+5 \\ L\left\{y\right\}=\frac{1}{{\left(p-3\right)}^2\left(p^2-6p+9\right)}+\frac{}{\left(p^2-6p+9\right)} \\ =\frac{1}{{\left(p-3\right)}^4}+\frac{5}{{\left(p-3\right)}^2} \end{gathered}$$

The inverse Laplace is,

$$\begin{gathered} y=L^{-1}\left\{\frac{1}{{\left(p-3\right)}^4}\right\}+L^{-1}\left\{\frac{5}{{\left(p-3\right)}^2}\right\} \\ =L^{-1}\left\{\frac{1}{{\left(o-3\right)}^4}\right\}+5L^{-1}\left\{\frac{1}{{\left(p-3\right)}^2}\right\} \\ =\frac{t^3}{3!}{e}^{3t}+5\left(\frac{t}{1!}{e}^{3t}\right)y \\ =e^{3x}\left(\frac{t^3}{6}+5t\right) \end{gathered}$$

Thus, the solution of given differential equation is$y=e^{3t}\left(\frac{t^3}{6}+5t\right)$.