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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y = sin (x), y^{^{'}} (\frac{π}{2}) = \frac{3}{2}, y (0) = 0

y^{^{''}} - 4 y^{^{'}} + 5 y = 125 t^{2}

\frac{d ^{3} y}{d x ^{3}} + \frac{d y}{d x} = tan (x)

3 x^{^{''}} + 25 x^{^{'}} + 48 x = 5 cos (4 t)

y^{^{''}} + 2 y^{^{'}} + 5 y = e^{- x} (2 x + sin (2 x))

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