What is the solution for y^{''}-y^'-2y=1-x,y(0)=1,y^'(0)=1 ?

The solution for y^{''}-y^'-2y=1-x,y(0)=1,y^'(0)=1 is y=(-1+7/4)e^{2x}+e^{-x}+x/2-3/4

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y^{''}-y^'-2y=1-x,y(0)=1,y^'(0)=1

y^{^{''}} - y^{^{'}} - 2 y = 1 - x, y (0) = 1, y^{^{'}} (0) = 1

y = (- 1 + \frac{7}{4}) e^{2 x} + e^{- x} + \frac{x}{2} - \frac{3}{4}

Solution steps

y^{''} (x) - y^{'} (x) - 2 y = 1 - x

Solve linear ODE:

y = (- 1 + \frac{7}{4}) e^{2 x} + e^{- x} + \frac{x}{2} - \frac{3}{4}

y = (- 1 + \frac{7}{4}) e^{2 x} + e^{- x} + \frac{x}{2} - \frac{3}{4}

y^{^{''}} + y = (t - 2 π), y (0) = 10, y^{^{'}} (0) = 0

y^{^{''}} + 16 y = 3.36

y^{^{''}} + 4 y^{^{'}} + 5 y = sin (x)

y^{^{''}} = 2 y + 20 tan^{3} (x), y (0) = 10 tan (x)

\frac{1}{6} y^{^{''}} + 6 y = 2 tan (6 t) - \frac{1}{4} e^{3 t}

What is the solution for y^{''}-y^'-2y=1-x,y(0)=1,y^'(0)=1 ?
The solution for y^{''}-y^'-2y=1-x,y(0)=1,y^'(0)=1 is y=(-1+7/4)e^{2x}+e^{-x}+x/2-3/4