What is the solution for y^{''}+4y=3cos(t) ?

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

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y^{''}+4y=3cos(t)

y^{^{''}} + 4 y = 3 cos (t)

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

Solution steps

y^{''} (t) + 4 y = 3 cos (t)

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} + 1 = 0

4 y^{^{''}} - y = \frac{e x}{2 + 4}

y^{^{''}} - 10 y^{^{'}} + 25 y = (\frac{e ^{- 5 x}}{1 + x ^{2}})

x^{^{''}} - 12 x^{^{'}} + 36 x = 2 t + e^{6 t}

y^{^{''}} + y = - 3, y (0) = 0, y^{^{'}} (0) = 1

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