What is the solution for y^{''}+4y=cos(t),y^'(0)=1,y(0)=1 ?

The solution for y^{''}+4y=cos(t),y^'(0)=1,y(0)=1 is y= 2/3 cos(2t)+1/2 sin(2t)+1/3 cos(t)

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y^{''}+4y=cos(t),y^'(0)=1,y(0)=1

y^{^{''}} + 4 y = cos (t), y^{^{'}} (0) = 1, y (0) = 1

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y = (x + 1) \cdot e^{x}

y^{^{''}} - 4 y = 16 x^{3}, y (0) = 2, y^{^{'}} (0) = - 14

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} = tan (2 x)

49 y^{^{''}} + 140 y^{^{'}} + 136 = 0

x^{^{''}} + \frac{1}{98} x = 8 cos (6 t)

What is the solution for y^{''}+4y=cos(t),y^'(0)=1,y(0)=1 ?
The solution for y^{''}+4y=cos(t),y^'(0)=1,y(0)=1 is y= 2/3 cos(2t)+1/2 sin(2t)+1/3 cos(t)