What is the solution for y^{''}+16y=3,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+16y=3,y(0)=0,y^'(0)=0 is y=-3/16 cos(4t)+3/16

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y^{''}+16y=3,y(0)=0,y^'(0)=0

y^{^{''}} + 16 y = 3, y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{3}{16} cos (4 t) + \frac{3}{16}

Solution steps

y^{''} (t) + 16 y = 3

Solve linear ODE:

y = - \frac{3}{16} cos (4 t) + \frac{3}{16}

y = - \frac{3}{16} cos (4 t) + \frac{3}{16}

y^{^{''}} + y^{^{'}} = x, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 12 y^{^{'}} + 36 y = 42 x + 2

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

y^{^{''}} + 2 y^{^{'}} - 3 y = e^{3 x} (2 + 8 x - 12 x^{2})

y^{^{''}} + 20 y^{^{'}} + 125 y = 9 sin (5 x)

What is the solution for y^{''}+16y=3,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+16y=3,y(0)=0,y^'(0)=0 is y=-3/16 cos(4t)+3/16