Find the solution to the initial value problem.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{6}\mathbf{t}\mathbf{;}\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{1}$

The differential equation is,

$\mathrm{y}\text{'}\text{'}=6\mathrm{t}\dots \left(1\right)$

Write the homogeneous differential equation of the equation (1),

${\mathrm{y}}^{\text{'}\text{'}}=0$

The auxiliary equation for the above equation,

${\mathrm{m}}^2=0$

The root of an auxiliary equation is,

${\mathrm{m}}_1=0,{\mathrm{m}}_2=0$

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1\mathrm{t}+{\mathrm{c}}_2$

Assume, the particular solution of equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{t}}^2(\mathrm{At}+\mathrm{B})......\left(2\right)\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{At}}^3+{\mathrm{Bt}}^2\end{array}$

Now find the first and second derivatives of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)=3{\mathrm{At}}^2+2\mathrm{Bt}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)=6\mathrm{At}+2\mathrm{B}\end{array}$

Substitute the value of ${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)$the equation (1),

$\begin{array}{c}\mathrm{y}\text{'}\text{'}=6\mathrm{t}\\ 6\mathrm{At}+2\mathrm{B}=6\mathrm{t}\end{array}$

Comparing all coefficients of the above equation,

$\begin{array}{c}6\mathrm{A}=6\Rightarrow \mathrm{A}=1\\ 2\mathrm{B}=0\Rightarrow \mathrm{B}=0\end{array}$

Substitute the value of A and B in the equation (2),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{t}}^2(\mathrm{At}+\mathrm{B})\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{t}}^2(\left(1\right)\mathrm{t}+0)\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{t}}^3\end{array}$

Therefore, the general solution is,

$\begin{array}{l}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)\\ \mathrm{y}={\mathrm{c}}_1\mathrm{t}+{\mathrm{c}}_2+{\mathrm{t}}^3......\left(3\right)\end{array}$

Given the initial condition,

$\mathrm{y}\left(0\right)=3,\mathrm{y}\text{'}\left(0\right)=-1$

Substitute the value of y = 3 and t = 0 in the equation (3),

$\begin{array}{c}\mathrm{y}={\mathrm{c}}_1\mathrm{t}+{\mathrm{c}}_2+{\mathrm{t}}^3\\ 3={\mathrm{c}}_1\left(0\right)+{\mathrm{c}}_2+0\\ {\mathrm{c}}_2=3\end{array}$

Now find the derivative of the equation (3),

$\mathrm{y}\text{'}={\mathrm{c}}_1+3{\mathrm{t}}^2$

Substitute the value of y’ = -1 and t = 0 in the above equation,

$\begin{array}{c}-1={\mathrm{c}}_1+3{\left(0\right)}^2\\ {\mathrm{c}}_1=-1\end{array}$

Substitute the value of${\mathrm{c}}_1=-1$ and${\mathrm{c}}_2=3$ in the equation (3), we get:

$\begin{array}{c}\mathrm{y}={\mathrm{c}}_1\mathrm{t}+{\mathrm{c}}_2+{\mathrm{t}}^3\\ \mathrm{y}=(-1)\mathrm{t}+3+{\mathrm{t}}^3\\ \mathrm{y}={\mathrm{t}}^3-\mathrm{t}+3\end{array}$

Thus, the general solution is$\mathrm{y}={\mathrm{t}}^3-\mathrm{t}+3.$