In Problems $\mathbf{1}\mathbf{-}\mathbf{14}$, solve the given initial value problem using the method of Laplace transforms

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{4}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{t}\mathbf{+}\mathbf{10}\mathbf{;}\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{3}$

• The Laplace transform is a strong integral transform used in mathematics to convert a function from the time domain to the s-domain.
• In some circumstances, the Laplace transform can be utilized to solve linear differential equations with given beginning conditions.
• $\mathbf{F}\left(\mathbf{s}\right)\mathbf{=}{\int }_{\mathbf{0}}^{\infty }\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}{\mathbf{e}}^{\mathbf{-}\mathbf{st}}\mathbf{t}\mathbf{\text{'}}$
  

Applying the Laplace transform and using its linearity we get

$$\begin{gathered} \mathcal{L}\left\{y\text{'}\text{'}+4y\right\}=\mathcal{L}\left\{4t^2-4t+10\right\} \\ \mathcal{L}\left\{y\text{'}\text{'}\right\}+4\mathcal{L}\left\{y\right\}=4\mathcal{L}\left\{t^2\right\}-4\mathcal{L}\left\{t\right\}+10\mathcal{L}\left\{1\right\} \end{gathered}$$

Solve for the transform as:

$$\begin{gathered} \left[s^{2}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)\right]+4Y\left(s\right)=\frac{8}{s^3}-\frac{4}{s^2}+\frac{10}{s} \\ {s}^{2}Y\left(s\right)+4Y\left(s\right)=\frac{8}{s^3}-\frac{4}{s^2}+\frac{10}{s}+3 \\ \left(s^2+4\right)Y\left(s\right)=\frac{3s^3+10s^2-4s+8}{s^3} \\ Y\left(s\right)=\frac{3s^3+10s^2-4s+8}{s^3\left(s^2+4\right)} \end{gathered}$$

Using partial fractions we get

$\frac{3s^3+10s^2-4s+8}{s^3\left(s^2+4\right)}=\frac{A}{s^3}+\frac{B}{s^2}+\frac{C}{s}+\frac{Ds+E}{s^2+4}$

Simplify the above relation as:

$$\begin{gathered} 3s^3+10s^2-4s+8=A\left(s^2-1\right)+B\left(s+5\right)\left(s-1\right)+C\left(s+5\right)\left(s+1\right) \\ =\left(C+D\right)s^4+\left(B+E\right)s^3+\left(A+4C\right)s^2+4Bs+4A \end{gathered}$$

Consider the obtained system of equations and their solution is:

$\left\{\begin{array}{l}\begin{array}{l}4A=8\\ 4B=-4\end{array}\\ B+E=3\\ C+D=0\end{array}\right.\Rightarrow \left\{\begin{array}{l}\begin{array}{l}A=2\\ B=-1\end{array}\\ C=2\\ D=-2\\ E=4\end{array}\right.$

Substitute the values and write the equation as:

$Y\left(s\right)=\frac{2}{s^3}-\frac{1}{s^2}+\frac{2}{s}+\frac{4-2s}{s^2+4}$

Using the inverse Laplace transform we obtain the solution of given differential equation.

$$\begin{gathered} y\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{2}{s^3}-\frac{1}{s^2}+\frac{2}{s}+\frac{4-2s}{s^2+4}\right\}\left(t\right) \\ =\mathcal{L}\left\{\frac{2}{s^3}\right\}-\mathcal{L}\left\{\frac{1}{s^2}\right\}+2\mathcal{L}\left\{\frac{1}{s}\right\}+2L\left\{\frac{2}{s^2+2^2}\right\}-2L\left\{\frac{2s}{s^2+2^2}\right\} \\ =t^2-t+2+2\sin 2t-2\cos 2t \end{gathered}$$

Therefore, $y\left(t\right)=t^2-t+2+2\sin 2t-2\cos 2t$

Therefore, the initial value for$y\text{'}\text{'}+4y=4t^2-4t+10$is$y\left(t\right)=t^2-t+2+2\sin 2t-2\cos 2t$