The differential equation we have is:

$y^{″}\left(t\right)+$ $y\left(t\right)=$ $6\theta \left(t-11\right)$

with initial conditions:

$y\left(0\right)=1$
$y^{\prime }\left(0\right)=$ $0$

This is a forced harmonic oscillator where the forcing turns on at $t=11$.

Solve the homogeneous part.

For $t\le 11$, we have $y^{″}\left(t\right)+$ $y\left(t\right)=$ $0$ with solution $y_h\left(t\right)=$ $c_1\cos \left(t\right)+$ $c_2\sin \left(t\right)$.

Considering the initial conditions $y_h\left(0\right)=$ $c_1=$ $1$ and ${y^{\prime }}_h\left(t\right)=$ $-c_1\sin \left(t\right)+$ $c_2\cos \left(t\right)$ so ${y^{\prime }}_h\left(t\right)=$ $c_2=$ $0$, we therefore have $y_h\left(t\right)=$ $\cos \left(t\right)$.

Solve the nonhomogeneous part.

For $t>11$, we have $y^{″}\left(t\right)+$ $y\left(t\right)=$ $6$ for which a particular solution is $y_p=6$, so:

$y_{\mathrm{nh}}\left(t\right)=$ $c_3\cos \left(t\right)+$ $c_4\sin \left(t\right)+$ $6$

Since the displacement and velocity must be continuous at $t=11$, from the homogeneous part we have $y\left(11\right)=$ $\cos \left(11\right)$ and $y^{\prime }\left(11\right)=$ $-\sin \left(11\right)$.

Therefore, $y_{\mathrm{nh}}\left(11\right)=$ $c_3\cos \left(11\right)+$ $c_4\sin \left(11\right)+$ $6=$ $\cos \left(11\right)$ and ${y_{\mathrm{nh}}}^{\prime }\left(11\right)=$ $-c_3\sin \left(11\right)+$ $c_4\cos \left(11\right)=$ $-\sin \left(11\right)$.

Thus, the constants $c_3$ and $c_4$ are:

$c_3=$ $1-$ $6\cos \left(11\right)$ and $c_4=$ $-6\sin \left(11\right)$

Therefore, $y_{\mathrm{nh}}\left(t\right)=$ $\left(1-6\cos \left(11\right)\right)\cos \left(t\right)-$ $\left(6\sin \left(11\right)\right)\sin \left(t\right)+$ $6$, which reduces to:

$y_{\mathrm{nh}}\left(t\right)=$ $6-$ $6\cos \left(11-t\right)+$ $\cos \left(t\right)$

The complete solution is:

$y\left(t\right)$ $=$ $\{\begin{array}{cc}\cos \left(t\right)& t\le 11\\ -6\cos \left(11-t\right)+\cos \left(t\right)+6& t>11\end{array}$

The corresponding plot is: