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

The solution for y^{''}+y^'-6y=3xe^x is y=c_{1}e^{2x}+c_{2}e^{-3x}-(3e^xx)/4-9/16 e^x

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y^{''}+y^'-6y=3xe^x

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

y = c_{1} e^{2 x} + c_{2} e^{- 3 x} - \frac{3 e ^{x} x}{4} - \frac{9}{16} e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 3 x} - \frac{3 e ^{x} x}{4} - \frac{9}{16} e^{x}

y = c_{1} e^{2 x} + c_{2} e^{- 3 x} - \frac{3 e ^{x} x}{4} - \frac{9}{16} e^{x}

3 y^{^{''}} + 27 y = 2 cos (3 x)

(3 D^{2} + 3) y = 3 x

y^{^{''}} + 3 y^{^{'}} + 2 y = 3 cos (2 t), y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} + y = 3 cos (t)

y^{^{''}} - 16 y = sin (2 x) - 3 cos (2 x)

What is the solution for y^{''}+y^'-6y=3xe^x ?
The solution for y^{''}+y^'-6y=3xe^x is y=c_{1}e^{2x}+c_{2}e^{-3x}-(3e^xx)/4-9/16 e^x