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

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

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

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

y = cos (2 t) - sin (2 t) + sin (t)

Solution steps

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

Solve linear ODE:

y = cos (2 t) - sin (2 t) + sin (t)

y = cos (2 t) - sin (2 t) + sin (t)

y^{^{''}} + 25 y = \frac{2}{3} cos (2) t

3 y^{^{''}} + y^{^{'}} = 9 x

x^{^{''}} + 5 x^{^{'}} + 6 x = 2 e^{- t}, x (0) = 1, x^{^{'}} (0) = 0

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

x^{^{''}} - x = 4 e^{t}

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