What is the solution for 4y^{''}+y=cos(t) ?

The solution for 4y^{''}+y=cos(t) is y=c_{1}cos(t/2)+c_{2}sin(t/2)-1/3 cos(t)

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4y^{''}+y=cos(t)

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y = 7 x sin (x)

y^{^{''}} - 2 y^{^{'}} = x + 1

y^{^{'''}} + y^{^{'}} = 3 + cos (x) - sin (x) + x

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

y^{^{''}} - 8 y^{^{'}} + 20 y = 200 x^{2} - 91 x e^{x}

What is the solution for 4y^{''}+y=cos(t) ?
The solution for 4y^{''}+y=cos(t) is y=c_{1}cos(t/2)+c_{2}sin(t/2)-1/3 cos(t)