What is the solution for x^{''}=-a^2x+b ?

The solution for x^{''}=-a^2x+b is x=c_{1}cos(at)+c_{2}sin(at)+b/(a^2)

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x^{''}=-a^2x+b

x^{^{''}} = - a^{2} x + b

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

Solution steps

x^{''} (t) = - a^{2} x + b

Solve linear ODE:

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

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

y^{^{''}} - 8 y^{^{'}} + 16 y = 2 e^{4 t} cos (5 t)

a^{2} y^{^{''}} - (a^{2} + 1) y^{^{'}} + y = x

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} - 12 y = 3 e^{5 x}

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} - 12 y = cos (x)

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

What is the solution for x^{''}=-a^2x+b ?
The solution for x^{''}=-a^2x+b is x=c_{1}cos(at)+c_{2}sin(at)+b/(a^2)