What is the solution for y^{''}+y^'-2y=(2+6x)e^x ?

The solution for y^{''}+y^'-2y=(2+6x)e^x is y=c_{1}e^x+c_{2}e^{-2x}+e^xx^2

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y^{''}+y^'-2y=(2+6x)e^x

y^{^{''}} + y^{^{'}} - 2 y = (2 + 6 x) e^{x}

y = c_{1} e^{x} + c_{2} e^{- 2 x} + e^{x} x^{2}

Solution steps

y^{''} (x) + y^{'} (x) - 2 y = (2 + 6 x) e^{x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 2 x} + e^{x} x^{2}

y = c_{1} e^{x} + c_{2} e^{- 2 x} + e^{x} x^{2}

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

y^{^{''}} + y = 9 cos (2 x)

y^{^{''}} - 12 y^{^{'}} + 36 y = 24 x + 2

y^{^{''}} + y = x^{2} + 1, y (0) = 5, y (1) = 0

y^{^{''}} + 25 y = 10 sin (5 t)

What is the solution for y^{''}+y^'-2y=(2+6x)e^x ?
The solution for y^{''}+y^'-2y=(2+6x)e^x is y=c_{1}e^x+c_{2}e^{-2x}+e^xx^2