What is the solution for x^{''}+ax=b ?

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

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x^{''}+ax=b

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 10 y^{^{'}} + 25 y = 3 e^{5 x}

y^{^{''}} + 6 y^{^{'}} + 9 y = 12 e^{t}

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 1, y (0) = 0, y^{^{'}} (0) = 0

D (x) = 82.5 - 2 \cdot 5 x

y^{^{''}} + 6 y = 8 t

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