What is the solution for y^{''}+9y=2x-5 ?

The solution for y^{''}+9y=2x-5 is y=c_{1}cos(3x)+c_{2}sin(3x)+(2x)/9-5/9

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y^{''}+9y=2x-5

y^{^{''}} + 9 y = 2 x - 5

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{2 x}{9} - \frac{5}{9}

Solution steps

y^{''} (x) + 9 y = 2 x - 5

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{2 x}{9} - \frac{5}{9}

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{2 x}{9} - \frac{5}{9}

(D^{2} + 1) v = 1

y^{^{''}} - y^{^{'}} + y = 5 sin (t)

x^{^{''}} + 16 x = 144, x (0) = 5, x^{^{'}} (0) = 12

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

x^{^{''}} + 6 x^{^{'}} + 9 = 0

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