What is the solution for y^{''}+9y=4+5x ?

The solution for y^{''}+9y=4+5x is y=c_{1}cos(3x)+c_{2}sin(3x)+(5x)/9+4/9

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y^{''}+9y=4+5x

y^{^{''}} + 9 y = 4 + 5 x

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{5 x}{9} + \frac{4}{9}

Solution steps

y^{''} (x) + 9 y = 4 + 5 x

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{5 x}{9} + \frac{4}{9}

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{5 x}{9} + \frac{4}{9}

y^{^{''}} - 3 y^{^{'}} = 15

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

y^{^{'''}} - y^{^{'}} = t e^{- t}

y^{^{''}} - y^{^{'}} = tan (x)

y^{^{''}} + 4 y = \frac{5}{2} cos (2 x)

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