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

The solution for y^{''}+36y=sin(6t) is y=c_{1}cos(6t)+c_{2}sin(6t)-(tcos(6t))/(12)

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

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

y = c_{1} cos (6 t) + c_{2} sin (6 t) - \frac{t cos ( 6 t )}{12}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (6 t) + c_{2} sin (6 t) - \frac{t cos ( 6 t )}{12}

y = c_{1} cos (6 t) + c_{2} sin (6 t) - \frac{t cos ( 6 t )}{12}

\int \frac{x ^{2} + 1}{x + 1} d x

sin (x) cos^{2} (x), 4 - 4 x, 0.69275, 0.92811

\frac{d y}{d x} + \frac{3 y}{x} + 2 = 3 x, y (1) = 1

\frac{d}{d x} (3 x^{2} + 5 x - 2)

\int 2 tan^{5} (4 x) d x

What is the solution for y^{''}+36y=sin(6t) ?
The solution for y^{''}+36y=sin(6t) is y=c_{1}cos(6t)+c_{2}sin(6t)-(tcos(6t))/(12)