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

The solution for y^{''}+16y=cos(3t),y(0)=0,y^'(0)=0 is y= 1/7 cos(3t)-1/7 cos(4t)

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y^{''}+16y=cos(3t),y(0)=0,y^'(0)=0

y^{^{''}} + 16 y = cos (3 t), y (0) = 0, y^{^{'}} (0) = 0

y = \frac{1}{7} cos (3 t) - \frac{1}{7} cos (4 t)

Solution steps

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

Solve linear ODE:

y = \frac{1}{7} cos (3 t) - \frac{1}{7} cos (4 t)

y = \frac{1}{7} cos (3 t) - \frac{1}{7} cos (4 t)

y^{^{''}} - y^{^{'}} - 6 y = - 14 t + 12 t^{2}

4 y^{^{''}} - 4 y^{^{'}} + y = 16 e \frac{t}{2}

(D^{2} - 2 D + 1) y = x e^{x} sin (x)

y^{^{''}} + 100 y = 200 sin (10 x), y^{^{'}} (0) = 0, y^{^{''}} (0) = 0

y^{^{''}} + 9 y = cos (3 x) + sin (3 x), y (0) = 2, y^{^{'}} (0) = 1

What is the solution for y^{''}+16y=cos(3t),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+16y=cos(3t),y(0)=0,y^'(0)=0 is y= 1/7 cos(3t)-1/7 cos(4t)