What is the solution for y^{''}+y=-sin(2t) ?

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

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y^{''}+y=-sin(2t)

y^{^{''}} + y = - sin (2 t)

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

Solution steps

y^{''} (t) + y = - sin (2 t)

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{5}{4 x - 3})

\int 2 (sin (x) - cos (x) + 1) d x

tan (5 x)

\int \frac{cos ( 4 x )}{2} d x

\int sec (x) tan (x) (\frac{x}{π}) d x

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