We are given the differential equation:

$4\frac{{d}^{2}y}{d{t}^{2}}+6\frac{dy}{dt}+5y=t$

with initial conditions $\frac{dy}{dt}(0)=y(0)=0$.

To find the Laplace transform of $y(t)$, we can apply the Laplace transform to both sides of the differential equation:

$4\mathcal{L}\left\{\frac{{d}^{2}y}{d{t}^{2}}\right\}+6\mathcal{L}\left\{\frac{dy}{dt}\right\}+5\mathcal{L}\{y\}=\mathcal{L}\{t\}$

Using the property of the Laplace transform that $\mathcal{L}\{\frac{{d}^{2}y}{d{t}^{2}}\}={s}^{2}\mathcal{L}\{y\}-s\xb7y(0)-\frac{dy}{dt}(0)$ and $\mathcal{L}\{\frac{dy}{dt}\}=s\mathcal{L}\{y\}-y(0)$, we can rewrite the equation as:

$4{s}^{2}Y(s)-4sy(0)-4\frac{dy}{dt}(0)+6sY(s)-6y(0)+5Y(s)=\frac{1}{{s}^{2}}$

Since $\frac{dy}{dt}(0)=\frac{d}{dt}(y(0))=0$ and $y(0)=0$, we can simplify the equation to:

$4{s}^{2}Y(s)+6sY(s)+5Y(s)=\frac{1}{{s}^{2}}$

Simplifying further, we get:

$(4{s}^{2}+6s+5)Y(s)=\frac{1}{{s}^{2}}$

Therefore, the Laplace transform of $y(t)$ is:

$Y(s)=\frac{1}{{s}^{2}(4{s}^{2}+6s+5)}$

To solve for $Y(s)$, we can use partial fraction decomposition. First, we factor the denominator:

$4{s}^{2}+6s+5=(2s+1{)}^{2}+4$

Thus, we can write:

$\frac{1}{{s}^{2}(4{s}^{2}+6s+5)}=\frac{As+B}{{s}^{2}}+\frac{Cs+D}{2s+1+\sqrt{4}}-\frac{Cs+D}{2s+1-\sqrt{4}}$

Simplifying, we get:

$Y(s)=\frac{1}{{s}^{2}}(\frac{As+B}{{s}^{2}}+\frac{Cs+D}{2s+1+2}-\frac{Cs+D}{2s+1-2})$

$Y(s)=\frac{As+B}{{s}^{4}}+\frac{Cs+D}{4{s}^{2}+4s}=\frac{As+B}{{s}^{4}}+\frac{Cs+D}{4s(s+1)}$

Using the initial condition $y(0)=0$, we know that $Y(0)=0$. Thus, we can solve for $B$:

$Y(0)=\frac{B}{{0}^{4}}+\frac{D}{4(0)(0+1)}=0\Rightarrow B=0$

Using the initial condition $\frac{dy}{dt}(0)=0$, we know that ${Y}^{\prime}(0)=0$. Thus, we can solve for $A$:

${Y}^{\prime}(s)=\frac{-2A{s}^{3}-4As+C{s}^{2}+D}{4{s}^{2}(s+1)}$

${Y}^{\prime}(0)=\frac{-0-0+C(0)+D}{4(0)(0+1)}=0\Rightarrow D=0$

Now we can write $Y(s)$ in terms of $A$ and $C$:

$Y(s)=\frac{As}{{s}^{4}}+\frac{C}{4(s+1)}$

To solve for $A$ and $C$, we can take the Laplace inverse of $Y(s)$ using tables or the method of partial fraction decomposition:

${\mathcal{L}}^{-1}\{Y(s)\}={\mathcal{L}}^{-1}\{\frac{As}{{s}^{4}}+\frac{C}{4(s+1)}\}$

$y(t)=a{t}^{3}+\frac{C}{4}{e}^{-t}$

where $a$ and $C$ are constants that we still need to solve for.

Using the initial condition $\frac{dy}{dt}(0)=0$, we can solve for $a$:

$\frac{dy}{dt}(t)=3a{t}^{2}-\frac{C}{4}{e}^{-t}$

$\frac{dy}{dt}(0)=0=3a(0{)}^{2}-\frac{C}{4}{e}^{-(0)}\Rightarrow C=0$

Thus, the solution to the differential equation with initial conditions is:

$y(t)=a{t}^{3}$

Using the initial condition $y(0)=0$, we get:

$y(0)=0=a(0{)}^{3}\Rightarrow a=0$

Therefore, the solution to the differential equation with initial conditions is $y(t)=0$.

In terms of Laplace transform, we have:

$Y(s)=\frac{1}{{s}^{2}(4{s}^{2}+6s+5)}=\frac{Cs+D}{4s(s+1)}=\frac{0}{{s}^{4}}$

where $C=0$ and $D=0$.