What is the solution for y^{''}+2y^'+y=g(t-1),y(0)=2,y^'(0)=3 ?

The solution for y^{''}+2y^'+y=g(t-1),y(0)=2,y^'(0)=3 is y=(2+3g)e^{-t}+(2g+5)te^{-t}+gt-3g

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y^{''}+2y^'+y=g(t-1),y(0)=2,y^'(0)=3

y^{^{''}} + 2 y^{^{'}} + y = g (t - 1), y (0) = 2, y^{^{'}} (0) = 3

y = (2 + 3 g) e^{- t} + (2 g + 5) t e^{- t} + g t - 3 g

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = g (t - 1)

Solve linear ODE:

y = (2 + 3 g) e^{- t} + (2 g + 5) t e^{- t} + g t - 3 g

y = (2 + 3 g) e^{- t} + (2 g + 5) t e^{- t} + g t - 3 g

y^{^{''}} + 4 y = t^{2} + 4 e^{t}

y^{^{''}} + 2 y = 12 sin (2 t), y (0) = 5

y^{^{''}} + y^{^{'}} - 2 y = 3 e^{- 2 t}, y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} - 2 y^{^{'}} + 2 y = 2 x, y (0) = 4, y^{^{'}} (0) = 8

y^{^{''}} - 25 y = e^{5 t}, y (0) = 2, y^{^{'}} (0) = 1

What is the solution for y^{''}+2y^'+y=g(t-1),y(0)=2,y^'(0)=3 ?
The solution for y^{''}+2y^'+y=g(t-1),y(0)=2,y^'(0)=3 is y=(2+3g)e^{-t}+(2g+5)te^{-t}+gt-3g