What is the solution for y^{''}+4y=4cos(2t),y(0)=0,y^'(0)=1 ?

The solution for y^{''}+4y=4cos(2t),y(0)=0,y^'(0)=1 is y=tsin(2t)+1/2 sin(2t)

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y^{''}+4y=4cos(2t),y(0)=0,y^'(0)=1

y^{^{''}} + 4 y = 4 cos (2 t), y (0) = 0, y^{^{'}} (0) = 1

y = t sin (2 t) + \frac{1}{2} sin (2 t)

Solution steps

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

Solve linear ODE:

y = t sin (2 t) + \frac{1}{2} sin (2 t)

y = t sin (2 t) + \frac{1}{2} sin (2 t)

What is the solution for y^{''}+4y=4cos(2t),y(0)=0,y^'(0)=1 ?
The solution for y^{''}+4y=4cos(2t),y(0)=0,y^'(0)=1 is y=tsin(2t)+1/2 sin(2t)