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

The solution for y^{''}+4y=g(t),y(0)=4,y^'(0)=0 is y=4cos(2t)-(gsin(2t))/8+(gt)/4

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

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

y = 4 cos (2 t) - \frac{g sin ( 2 t )}{8} + \frac{g t}{4}

Solution steps

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

Solve linear ODE:

y = 4 cos (2 t) - \frac{g sin ( 2 t )}{8} + \frac{g t}{4}

y = 4 cos (2 t) - \frac{g sin ( 2 t )}{8} + \frac{g t}{4}

y^{^{''}} - 3 + y^{^{'}} + 2 y = 0

y^{^{''}} - 4 y = 4 x

x^{^{''}} + 2 x^{^{'}} + 151 x = 8 cos (12 t)

y^{^{''}} + 3 y = - 48 t^{2} e^{3} t

y^{^{''''}} - 256 y = 2 e^{- 4 x} + 3 e^{3 x} + cos (4 x) + 2

What is the solution for y^{''}+4y=g(t),y(0)=4,y^'(0)=0 ?
The solution for y^{''}+4y=g(t),y(0)=4,y^'(0)=0 is y=4cos(2t)-(gsin(2t))/8+(gt)/4