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

The solution for y^{''}+2y^'=1+(t-2),y(0)=0,y^'(0)=1 is y= 7/8-7/8 e^{-2t}+(t^2)/4-(3t)/4

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

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

y = \frac{7}{8} - \frac{7}{8} e^{- 2 t} + \frac{t ^{2}}{4} - \frac{3 t}{4}

Solution steps

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

Solve linear ODE:

y = \frac{7}{8} - \frac{7}{8} e^{- 2 t} + \frac{t ^{2}}{4} - \frac{3 t}{4}

y = \frac{7}{8} - \frac{7}{8} e^{- 2 t} + \frac{t ^{2}}{4} - \frac{3 t}{4}

What is the solution for y^{''}+2y^'=1+(t-2),y(0)=0,y^'(0)=1 ?
The solution for y^{''}+2y^'=1+(t-2),y(0)=0,y^'(0)=1 is y= 7/8-7/8 e^{-2t}+(t^2)/4-(3t)/4