What is the solution for y^{''}+16y=6sin(4t),y(0)=2,y^'(0)=0 ?

The solution for y^{''}+16y=6sin(4t),y(0)=2,y^'(0)=0 is y=2cos(4t)+3/16 sin(4t)-(3tcos(4t))/4

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y^{''}+16y=6sin(4t),y(0)=2,y^'(0)=0

y^{^{''}} + 16 y = 6 sin (4 t), y (0) = 2, y^{^{'}} (0) = 0

y = 2 cos (4 t) + \frac{3}{16} sin (4 t) - \frac{3 t cos ( 4 t )}{4}

Solution steps

y^{''} (t) + 16 y = 6 sin (4 t)

Solve linear ODE:

y = 2 cos (4 t) + \frac{3}{16} sin (4 t) - \frac{3 t cos ( 4 t )}{4}

y = 2 cos (4 t) + \frac{3}{16} sin (4 t) - \frac{3 t cos ( 4 t )}{4}

y^{^{''}} - 5 y^{^{'}} + 6 y = 6 x + 2 - 3 e^{2}

y^{^{''}} + 2 y^{^{'}} - 3 y = - 20 sin (x)

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

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{2 t} ln (t)

y^{^{''}} - 7 y^{^{'}} + 6 y = x, y^{1} (0) = e x

What is the solution for y^{''}+16y=6sin(4t),y(0)=2,y^'(0)=0 ?
The solution for y^{''}+16y=6sin(4t),y(0)=2,y^'(0)=0 is y=2cos(4t)+3/16 sin(4t)-(3tcos(4t))/4