What is the solution for y^{''}+36y=-24sin(6t) ?

The solution for y^{''}+36y=-24sin(6t) is y=c_{1}cos(6t)+c_{2}sin(6t)+2tcos(6t)

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y^{''}+36y=-24sin(6t)

y^{^{''}} + 36 y = - 24 sin (6 t)

y = c_{1} cos (6 t) + c_{2} sin (6 t) + 2 t cos (6 t)

Solution steps

y^{''} (t) + 36 y = - 24 sin (6 t)

Solve linear ODE:

y = c_{1} cos (6 t) + c_{2} sin (6 t) + 2 t cos (6 t)

y = c_{1} cos (6 t) + c_{2} sin (6 t) + 2 t cos (6 t)

(x e^{x})^{^{'}}

(e^{x} + y) d x + (e^{y} + x) d y = 0

\frac{\partial}{\partial x} (2 x sin (y))

\frac{d}{d x} ((x - 1)^{\frac{1}{2}})

\frac{d}{d x} (sin (3 x^{2}) + 2 x^{3})

What is the solution for y^{''}+36y=-24sin(6t) ?
The solution for y^{''}+36y=-24sin(6t) is y=c_{1}cos(6t)+c_{2}sin(6t)+2tcos(6t)