What is the solution for 16y^{''}-24y^'+9y=5-3x ?

The solution for 16y^{''}-24y^'+9y=5-3x is y=c_{1}e^{(3x)/4}+c_{2}e^{(3x)/4}x-x/3-1/3

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16y^{''}-24y^'+9y=5-3x

16 y^{^{''}} - 24 y^{^{'}} + 9 y = 5 - 3 x

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

Solution steps

16 y^{''} (x) - 24 y^{'} (x) + 9 y = 5 - 3 x

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{- 2 x} \cdot x^{- 2}

y^{^{''}} + 4 y^{^{'}} + 40 y = 50 sin (6 t)

y^{^{''}} - 4 y + 3 y = 2 x + 1

y^{^{''}} + 4 y = sin (3 t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 6 y^{^{'}} - 7 y = e^{- x}

What is the solution for 16y^{''}-24y^'+9y=5-3x ?
The solution for 16y^{''}-24y^'+9y=5-3x is y=c_{1}e^{(3x)/4}+c_{2}e^{(3x)/4}x-x/3-1/3