What is the solution for y^{''}+b^2y=b ?

The solution for y^{''}+b^2y=b is y=c_{1}cos(bt)+c_{2}sin(bt)+1/b

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y^{''}+b^2y=b

y^{^{''}} + b^{2} y = b

y = c_{1} cos (b t) + c_{2} sin (b t) + \frac{1}{b}

Solution steps

y^{''} (t) + b^{2} y = b

Solve linear ODE:

y = c_{1} cos (b t) + c_{2} sin (b t) + \frac{1}{b}

y = c_{1} cos (b t) + c_{2} sin (b t) + \frac{1}{b}

y^{^{'''}} + 225 y^{^{'}} = tan (15 x)

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

y^{^{''}} - 12 y^{^{'}} + 36 y = 12 x + 3

y^{^{''}} + 25 y = 3 sec (5 x)

y^{^{''}} - 4 y^{^{'}} + 20 y = 3 x^{2} - 1

What is the solution for y^{''}+b^2y=b ?
The solution for y^{''}+b^2y=b is y=c_{1}cos(bt)+c_{2}sin(bt)+1/b