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

The solution for y^{''}+4y=3sin(2t),y(0)=2,y^'(0)=-1 is y=2cos(2t)-1/8 sin(2t)-(3tcos(2t))/4

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y^{''}+4y=3sin(2t),y(0)=2,y^'(0)=-1

y^{^{''}} + 4 y = 3 sin (2 t), y (0) = 2, y^{^{'}} (0) = - 1

y = 2 cos (2 t) - \frac{1}{8} sin (2 t) - \frac{3 t cos ( 2 t )}{4}

Solution steps

y^{''} (t) + 4 y = 3 sin (2 t)

Solve linear ODE:

y = 2 cos (2 t) - \frac{1}{8} sin (2 t) - \frac{3 t cos ( 2 t )}{4}

y = 2 cos (2 t) - \frac{1}{8} sin (2 t) - \frac{3 t cos ( 2 t )}{4}

x^{^{''}} + 9 x = 5 sin (3 t)

y^{^{''}} - 4 y = (x^{2} - 7) sin (2 x)

y^{^{''}} - 36 y = sinh (6 x)

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

y^{^{''}} - y^{^{'}} + y = \frac{sin ( t )}{3} - 2 e^{2 t}

What is the solution for y^{''}+4y=3sin(2t),y(0)=2,y^'(0)=-1 ?
The solution for y^{''}+4y=3sin(2t),y(0)=2,y^'(0)=-1 is y=2cos(2t)-1/8 sin(2t)-(3tcos(2t))/4