What is the solution for y^{''}+3y=-48xe^{3x} ?

The solution for y^{''}+3y=-48xe^{3x} is y=c_{1}cos(sqrt(3)x)+c_{2}sin(sqrt(3)x)-4xe^{3x}+2e^{3x}

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y^{''}+3y=-48xe^{3x}

y^{^{''}} + 3 y = - 48 x e^{3 x}

y = c_{1} cos (3 x) + c_{2} sin (3 x) - 4 x e^{3 x} + 2 e^{3 x}

Solution steps

y^{''} (x) + 3 y = - 48 x e^{3 x}

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) - 4 x e^{3 x} + 2 e^{3 x}

y = c_{1} cos (3 x) + c_{2} sin (3 x) - 4 x e^{3 x} + 2 e^{3 x}

y^{^{''}} + 8 y^{^{'}} + 8 y = cos (4 x)

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

x^{^{''}} + 6 x^{^{'}} + 34 x = 68

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 x^{2} - 1

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

What is the solution for y^{''}+3y=-48xe^{3x} ?
The solution for y^{''}+3y=-48xe^{3x} is y=c_{1}cos(sqrt(3)x)+c_{2}sin(sqrt(3)x)-4xe^{3x}+2e^{3x}