What is the solution for y^{''}+9y=xe^{3x} ?

The solution for y^{''}+9y=xe^{3x} is y=c_{1}cos(3x)+c_{2}sin(3x)+(e^{3x}x)/(18)-1/54 e^{3x}

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y^{''}+9y=xe^{3x}

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

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{e ^{3 x} x}{18} - \frac{1}{54} e^{3 x}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{e ^{3 x} x}{18} - \frac{1}{54} e^{3 x}

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{e ^{3 x} x}{18} - \frac{1}{54} e^{3 x}

(D^{2} - 3 D + 2) y = 72 x e^{- x}

y^{^{''}} + 6 y^{^{'}} + 8 = (3 x - 2) e^{4 x}

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{- 4 t}, y (0) = 1, y (0) = 5

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

\frac{d ^{2} y}{d x ^{2}} + 9 y = sin (3 x)

What is the solution for y^{''}+9y=xe^{3x} ?
The solution for y^{''}+9y=xe^{3x} is y=c_{1}cos(3x)+c_{2}sin(3x)+(e^{3x}x)/(18)-1/54 e^{3x}