What is the solution for y^{''}+36y=-48sin(6t) ?

The solution for y^{''}+36y=-48sin(6t) is y=c_{1}cos(6t)+c_{2}sin(6t)+4tcos(6t)

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y^{''}+36y=-48sin(6t)

y^{^{''}} + 36 y = - 48 sin (6 t)

y = c_{1} cos (6 t) + c_{2} sin (6 t) + 4 t cos (6 t)

Solution steps

y^{''} (t) + 36 y = - 48 sin (6 t)

Solve linear ODE:

y = c_{1} cos (6 t) + c_{2} sin (6 t) + 4 t cos (6 t)

y = c_{1} cos (6 t) + c_{2} sin (6 t) + 4 t cos (6 t)

\frac{d}{d x} (x^{2} + x - 6)

f (x) = (3 x^{2} - 4 x) e^{x}

(x - x) (x + x)

x \to 0 lim ((1 - \frac{3}{x})^{x})

\int \frac{1}{r ^{2}} d r

What is the solution for y^{''}+36y=-48sin(6t) ?
The solution for y^{''}+36y=-48sin(6t) is y=c_{1}cos(6t)+c_{2}sin(6t)+4tcos(6t)