What is the solution for y^{''}+16y=g(t),y(0)=0,y^'(0)=2 ?

The solution for y^{''}+16y=g(t),y(0)=0,y^'(0)=2 is y=(gt)/(16)+(sin(4t)(32-g))/(64)

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y^{''}+16y=g(t),y(0)=0,y^'(0)=2

y^{^{''}} + 16 y = g (t), y (0) = 0, y^{^{'}} (0) = 2

y = \frac{g t}{16} + \frac{sin ( 4 t ) ( 32 - g )}{64}

Solution steps

y^{''} (t) + 16 y = g (t)

Solve linear ODE:

y = \frac{g t}{16} + \frac{sin ( 4 t ) ( 32 - g )}{64}

y = \frac{g t}{16} + \frac{sin ( 4 t ) ( 32 - g )}{64}

y^{^{''}} - 2 y^{^{'}} + y = 3 x^{\frac{3}{2}} e^{x}

y^{^{''}} + 16 y = g (t), y (0) = 0, y^{^{'}} (0) = 5

2 y^{^{''}} + 5 y^{^{'}} + 2 y = x e^{- x}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 2 y^{^{'}} - 3 y = 2 e^{x}

y^{^{''}} + 4 y^{^{'}} + 5 y = 2 e^{- 2 x} cos (x)

What is the solution for y^{''}+16y=g(t),y(0)=0,y^'(0)=2 ?
The solution for y^{''}+16y=g(t),y(0)=0,y^'(0)=2 is y=(gt)/(16)+(sin(4t)(32-g))/(64)