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

The solution for y^{''}+2y^'+y=t,y(0)=1,y^'(0)=2 is y=3e^{-t}+4te^{-t}+t-2

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

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

y = 3 e^{- t} + 4 t e^{- t} + t - 2

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = t

Solve linear ODE:

y = 3 e^{- t} + 4 t e^{- t} + t - 2

y = 3 e^{- t} + 4 t e^{- t} + t - 2

y^{^{''}} - 3 y^{^{'}} - 10 y = 6 cos (2 t)

y^{^{''}} - 2 y^{^{'}} + y = x^{- 1} e^{x}

(d^{2} - 2 d + 2) y = 3 e^{x} sin (x)

y^{^{''}} + 5 y^{^{'}} + 6 y = 5 e^{3 t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = 2 e^{2 x}

What is the solution for y^{''}+2y^'+y=t,y(0)=1,y^'(0)=2 ?
The solution for y^{''}+2y^'+y=t,y(0)=1,y^'(0)=2 is y=3e^{-t}+4te^{-t}+t-2