What is the solution for y^{''}-y^'-6y=12x-e^x,y(0)=1 ?

The solution for y^{''}-y^'-6y=12x-e^x,y(0)=1 is y=(-c_{2}+1/2)e^{3x}+c_{2}e^{-2x}-2x+1/3+1/6 e^x

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y^{''}-y^'-6y=12x-e^x,y(0)=1

y^{^{''}} - y^{^{'}} - 6 y = 12 x - e^{x}, y (0) = 1

y = (- c_{2} + \frac{1}{2}) e^{3 x} + c_{2} e^{- 2 x} - 2 x + \frac{1}{3} + \frac{1}{6} e^{x}

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = 12 x - e^{x}

Solve linear ODE:

y = (- c_{2} + \frac{1}{2}) e^{3 x} + c_{2} e^{- 2 x} - 2 x + \frac{1}{3} + \frac{1}{6} e^{x}

y = (- c_{2} + \frac{1}{2}) e^{3 x} + c_{2} e^{- 2 x} - 2 x + \frac{1}{3} + \frac{1}{6} e^{x}

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = 2 x

\frac{d ^{2} y}{d x ^{2}} + y = sec (x) tan (x)

y^{^{''}} + y + y = \frac{( e ^{t} )}{t}

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

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

What is the solution for y^{''}-y^'-6y=12x-e^x,y(0)=1 ?
The solution for y^{''}-y^'-6y=12x-e^x,y(0)=1 is y=(-c_{2}+1/2)e^{3x}+c_{2}e^{-2x}-2x+1/3+1/6 e^x