$y"+3y'+2y=g(t), y(0)=0, y'(0)=-2,$where,
$g\left( t \right) = \left\{ {\begin{array}{*{20}{l}}2&{\hspace{0.25in}t < 6}\\t&{\hspace{0.25in}6 \le t < 10}\\4&{\hspace{0.25in}t \ge 10}\end{array}} \right.$
Now, Author got solutions as follows $f(t)=\frac12-e^{-t}+\frac12e^{-2t}$
$g(t)=-\frac34+\frac12 t+e^{-t}-\frac14 e^{-2t}$
$h(t)=2e^{-t}-2e^{-2t}$.
Now the author says If we had not used Laplace transform, For 1st part of g(t), we get solution $y_1(t)=2f(t)-h(t)$
For 2nd part of g(t), we get solution $y_2(t)=2f(t)-h(t)+4f(t-6)+g(t-6)$
For 3rd part of g(t), we get solution $y_3(t)=2f(t)-h(t)+4f(t-6)+g(t-6)-6f(t-10)-g(t-10)$ Now i want to verify $y_2(t)$ when t=7, How can i verify it?
What would be the initial conditions? Are they $y(7)=y_1(7)$ and $y'(7)=y'_1(7)$
Note that if you need any additional information, it will be provided.
No, you need to have the differentiable phase change at $t=6$. That is $$ y_2(6)=y_1(6)\\ y_2'(6)=y_1'(6) $$ To understand the components, check that, $L$ being the differential operator on the left side, $L(y)=y''+3y'+2y$, then \begin{align} L(f)&=1&f(0)&=0&f'(0)&=0\\ L(g)&=t&g(0)&=0&g'(0)&=0\\ L(h)&=0&h(0)&=0&h'(0)&=2\\ \end{align} so that the first two provide the components for the right side, and $h$ accounts for the initial condition.