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

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

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

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

y = c_{1} e^{- t} + c_{2} t e^{- t} + t - 2

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} t e^{- t} + t - 2

y = c_{1} e^{- t} + c_{2} t e^{- t} + t - 2

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

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

y^{^{''''}} + 9 y^{^{''}} = sec^{2} (3 t)

y^{^{''}} - 8 y^{^{'}} + 16 y = 8 x + 7

y^{^{''}} - 6 y^{^{'}} - 7 y = - 9 e^{- 2 x}

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