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}

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