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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + y = e^{- x}, y^{^{'}} (0) = 0

(D^{4} - 4 D^{2}) y = 5 x^{2} - e^{2 x}

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

y^{^{''}} + 25 y = 50 sin (5 t)

y^{^{''}} + 2 y^{^{'}} + 5 y = cos (2 x) - 4 sin (2 x)

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