What is the solution for y^{''}+3y=sin(t) ?

The solution for y^{''}+3y=sin(t) is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)+1/2 sin(t)

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y^{''}+3y=sin(t)

y^{^{''}} + 3 y = sin (t)

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{1}{2} sin (t)

Solution steps

y^{''} (t) + 3 y = sin (t)

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{1}{2} sin (t)

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{1}{2} sin (t)

f (x) = ln (x + x^{2} - 1)

\int (6 x - 5) d x

\frac{d}{d x} (lo g_{e} (cos (x)))

\frac{d}{d x} (\frac{1 + 2 x}{( 3 x - 1 ) ^{2}})

(y + 4) \frac{d y}{d x} = 2 x y + 4 y

What is the solution for y^{''}+3y=sin(t) ?
The solution for y^{''}+3y=sin(t) is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)+1/2 sin(t)