What is the solution for y^{''}+y=\H(t)-2\H(t-b/2)+\H(t-b) ?

The solution for y^{''}+y=\H(t)-2\H(t-b/2)+\H(t-b) is

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y^{''}+y=\H(t)-2\H(t-b/2)+\H(t-b)

y^{^{''}} + y = H (t) - 2 H (t - \frac{b}{2}) + H (t - b)

y = c_{1} cos (t) + c_{2} sin (t) + cos (t) \cdot \int - sin (t) (H (t) - 2 H (t - \frac{b}{2}) + H (t - b)) d t + sin (t) \cdot \int cos (t) (H (t) - 2 H (t - \frac{b}{2}) + H (t - b)) d t

Solution steps

y^{''} (t) + y = H (t) - 2 H (t - \frac{b}{2}) + H (t - b)

Solve linear ODE:

y = c_{1} cos (t) + c_{2} sin (t) + cos (t) \cdot \int - sin (t) (H (t) - 2 H (t - \frac{b}{2}) + H (t - b)) d t + sin (t) \cdot \int cos (t) (H (t) - 2 H (t - \frac{b}{2}) + H (t - b)) d t

y = c_{1} cos (t) + c_{2} sin (t) + cos (t) \cdot \int - sin (t) (H (t) - 2 H (t - \frac{b}{2}) + H (t - b)) d t + sin (t) \cdot \int cos (t) (H (t) - 2 H (t - \frac{b}{2}) + H (t - b)) d t

What is the solution for y^{''}+y=\H(t)-2\H(t-b/2)+\H(t-b) ?
The solution for y^{''}+y=\H(t)-2\H(t-b/2)+\H(t-b) is