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

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

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

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

y = 2 cos (2 t) + \frac{1}{2} sin (2 t) + \frac{3 t sin ( 2 t )}{2}

Solution steps

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

Solve linear ODE:

y = 2 cos (2 t) + \frac{1}{2} sin (2 t) + \frac{3 t sin ( 2 t )}{2}

y = 2 cos (2 t) + \frac{1}{2} sin (2 t) + \frac{3 t sin ( 2 t )}{2}

y^{^{''}} - 9 y = - 27 t + 18 t^{2}

y^{^{'''}} - 2 y^{^{''}} + 26 y^{^{'}} = 174 e^{3 x}

y^{^{''}} + y^{^{'}} - y = t^{3}, y (0) = 1, y^{^{'}} (0) = 0

- y^{^{''}} + 2 y^{^{'}} + 3 y = 1, y (0) = 0, y (1) = (0)

y^{^{''}} - 3 y = e^{- 2 x}

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