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

The solution for y^{''}+4y=cos(2t),y(0)=0,y^'(0)=0 is y=(tsin(2t))/4

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y^{''}+4y=cos(2t),y(0)=0,y^'(0)=0

y^{^{''}} + 4 y = cos (2 t), y (0) = 0, y^{^{'}} (0) = 0

y = \frac{t sin ( 2 t )}{4}

Solution steps

y^{''} (t) + 4 y = cos (2 t)

Solve linear ODE:

y = \frac{t sin ( 2 t )}{4}

y = \frac{t sin ( 2 t )}{4}

y^{^{''}} + y^{^{'}} + 5 y = 4 e^{- t} cos (2 t)

y^{^{''}} - 5 y^{^{'}} + 4 y = e^{2 t}

y^{^{''}} + 6 y^{^{'}} + 8 y = 19 t e^{2 t}

y^{^{''}} - 6 y^{^{'}} + 8 y = 4 e^{4 x}

y^{^{''}} - y = 10 sin^{2} (x)

What is the solution for y^{''}+4y=cos(2t),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+4y=cos(2t),y(0)=0,y^'(0)=0 is y=(tsin(2t))/4