What is the solution for y^{''}+4y=g(t),y(0)=3,y^'(0)=0 ?

The solution for y^{''}+4y=g(t),y(0)=3,y^'(0)=0 is y=3cos(2t)-(gsin(2t))/8+(gt)/4

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y^{''}+4y=g(t),y(0)=3,y^'(0)=0

y^{^{''}} + 4 y = g (t), y (0) = 3, y^{^{'}} (0) = 0

y = 3 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 = 3 cos (2 t) - \frac{g sin ( 2 t )}{8} + \frac{g t}{4}

y = 3 cos (2 t) - \frac{g sin ( 2 t )}{8} + \frac{g t}{4}

y^{^{''}} - 12 y^{^{'}} + 36 y = 24 x + 6

y^{^{''}} - 12 y^{^{'}} + 36 y = 24 x + 5

y^{^{''}} - 4 y^{^{'}} + 4 y = 6 x + 4

y^{^{''}} + 4 y^{^{'}} + 3 y = 2 e^{- 2 t}

y^{^{''}} - 10 y^{^{'}} + 16 y = 17 sin (2 x)

What is the solution for y^{''}+4y=g(t),y(0)=3,y^'(0)=0 ?
The solution for y^{''}+4y=g(t),y(0)=3,y^'(0)=0 is y=3cos(2t)-(gsin(2t))/8+(gt)/4