What is the solution for y^{''}+4y=6sin(2t),y(0)=2,y^'(0)=1 ?

The solution for y^{''}+4y=6sin(2t),y(0)=2,y^'(0)=1 is y=2cos(2t)+5/4 sin(2t)-(3tcos(2t))/2

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y^{''}+4y=6sin(2t),y(0)=2,y^'(0)=1

y^{^{''}} + 4 y = 6 sin (2 t), y (0) = 2, y^{^{'}} (0) = 1

y = 2 cos (2 t) + \frac{5}{4} sin (2 t) - \frac{3 t cos ( 2 t )}{2}

Solution steps

y^{''} (t) + 4 y = 6 sin (2 t)

Solve linear ODE:

y = 2 cos (2 t) + \frac{5}{4} sin (2 t) - \frac{3 t cos ( 2 t )}{2}

y = 2 cos (2 t) + \frac{5}{4} sin (2 t) - \frac{3 t cos ( 2 t )}{2}

What is the solution for y^{''}+4y=6sin(2t),y(0)=2,y^'(0)=1 ?
The solution for y^{''}+4y=6sin(2t),y(0)=2,y^'(0)=1 is y=2cos(2t)+5/4 sin(2t)-(3tcos(2t))/2