What is the solution for y^{''}+100y=3cos(8t),y(0)=0,y^'(0)=0 ?

The solution for y^{''}+100y=3cos(8t),y(0)=0,y^'(0)=0 is y= 1/12 cos(8t)-1/12 cos(10t)

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y^{''}+100y=3cos(8t),y(0)=0,y^'(0)=0

y^{^{''}} + 100 y = 3 cos (8 t), y (0) = 0, y^{^{'}} (0) = 0

y = \frac{1}{12} cos (8 t) - \frac{1}{12} cos (10 t)

Solution steps

y^{''} (t) + 100 y = 3 cos (8 t)

Solve linear ODE:

y = \frac{1}{12} cos (8 t) - \frac{1}{12} cos (10 t)

y = \frac{1}{12} cos (8 t) - \frac{1}{12} cos (10 t)

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

y^{^{''}} - 4 y^{^{'}} + 4 y = - 3 + 2 x + 2 x^{2}

y^{^{''}} - 9 y = sin (2 x)

\frac{d ^{2} y}{d x ^{2}} + y = 32 x^{3}

y^{^{''}} - 0.16 y = 32 cosh (0.4 x)

What is the solution for y^{''}+100y=3cos(8t),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+100y=3cos(8t),y(0)=0,y^'(0)=0 is y= 1/12 cos(8t)-1/12 cos(10t)