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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to 6 lim (8 x^{2} - 3 x + 1)

x \to - 5 + lim (\frac{5 x}{x + 5})

f (x) = 2 x^{2} - 9 x + 10

f^{^{'}} (x) = 6 x + 1 2^{\frac{3}{4}}

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