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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x^{^{''}} - 2 x^{^{'}} - 3 x = e^{4 t}

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

y^{^{'''}} - 6 y^{^{''}} = x^{2}

y^{^{'''}} - 6 y^{^{'}} = 3 - cos (x)

y^{^{'''}} - 3 y^{^{''}} = 9 e^{x}

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