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

The solution for y^{''}+ay=b is y=c_{1}cos(sqrt(a)t)+c_{2}sin(sqrt(a)t)+b/a

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y^{''}+ay=b

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} + 25 y = 3 x, y (0) = 1, y^{^{'}} (0) = 5

y^{^{'''}} + 7 y^{^{''}} + 15 y^{^{'}} + 9 = 1

y^{^{'''}} - y^{^{'}} = t e^{- t} + 2 cos (t)

x^{^{''}} + 9 x = 8 sin (3 t)

What is the solution for y^{''}+ay=b ?
The solution for y^{''}+ay=b is y=c_{1}cos(sqrt(a)t)+c_{2}sin(sqrt(a)t)+b/a