What is the solution for y^{''}+y^'-2y+1=0 ?

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

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y^{''}+y^'-2y+1=0

y^{^{''}} + y^{^{'}} - 2 y + 1 = 0

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} - 12 = 3 cos (x)

y^{^{''}} + 4 y = - 1

y^{^{''}} - 3 y^{^{'}} + 2 y = \frac{e ^{2 x}}{1 + e ^{2 x}}

(d^{2} - 2 d + 2) y = 4 x - 2 + 2 e^{x} sin (x)

y^{^{''}} - 12 y^{^{'}} + 36 y = 36 x + 4

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