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

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

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

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

y = (- c_{2} + \frac{1}{2}) e^{t} + c_{2} e^{- 2 t} - t - \frac{1}{2}

Solution steps

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

Solve linear ODE:

y = (- c_{2} + \frac{1}{2}) e^{t} + c_{2} e^{- 2 t} - t - \frac{1}{2}

y = (- c_{2} + \frac{1}{2}) e^{t} + c_{2} e^{- 2 t} - t - \frac{1}{2}

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

y^{^{''}} + 3 y^{^{'}} + 2 y = 4 e^{x}

2 y^{^{''}} + 4 y^{^{'}} + 2 y = e^{2 x} cos (2 x)

x^{^{''}} - x = 1

2 y^{^{''}} - 4 y^{^{'}} + 4 y = e^{x} tan (x) sec (x)

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