What is the solution for y^{''}+y^'-6y=2e^{-3x}+e^{2x} ?

The solution for y^{''}+y^'-6y=2e^{-3x}+e^{2x} is y=c_{1}e^{2x}+c_{2}e^{-3x}+(-2e^{-3x}x+e^{2x}x)/5

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y^{''}+y^'-6y=2e^{-3x}+e^{2x}

y^{^{''}} + y^{^{'}} - 6 y = 2 e^{- 3 x} + e^{2 x}

y = c_{1} e^{2 x} + c_{2} e^{- 3 x} + \frac{- 2 e ^{- 3 x} x + e ^{2 x} x}{5}

Solution steps

y^{''} (x) + y^{'} (x) - 6 y = 2 e^{- 3 x} + e^{2 x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 3 x} + \frac{- 2 e ^{- 3 x} x + e ^{2 x} x}{5}

y = c_{1} e^{2 x} + c_{2} e^{- 3 x} + \frac{- 2 e ^{- 3 x} x + e ^{2 x} x}{5}

y^{^{''}} + 4 y = - cos (3 t) - 2 sin (3 t)

y^{^{'''}} - 4 y^{^{''}} + 5 y^{^{'}} = 40 x - 7 - 6 e^{2 x}

y^{^{'''}} = 3 y^{^{''}} + 9 (e^{9 x}) + 9 (cos (x))

y^{^{''}} - 5 y^{^{'}} + 6 y = 8 e^{4 x}

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 3 y = 20 cos (x)

What is the solution for y^{''}+y^'-6y=2e^{-3x}+e^{2x} ?
The solution for y^{''}+y^'-6y=2e^{-3x}+e^{2x} is y=c_{1}e^{2x}+c_{2}e^{-3x}+(-2e^{-3x}x+e^{2x}x)/5