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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y = x^{2} + 1, y (0) = 4, y (1) = 0

y^{^{''}} + y = tan (x), y (0) = 0, y^{^{'}} (0) = 1

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

x^{^{''}} - 5 x^{^{'}} + 6 x = t^{2} e^{3 t}

\frac{d ^{2} y}{d x ^{2}} - 9 y = sin (3 x)

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