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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y = 3 e^{2 x}, y (0) = 0, y^{^{'}} (0) = 0

2 y^{^{''}} + 3 y^{^{'}} - 2 y = t e^{- 2 t}

x^{^{''}} + 0.8 x^{^{'}} + 19.6 x = 4 e^{- 2 t} cos (6 t)

y^{^{''}} + 3 y^{^{'}} + 2 y = u_{2} (t), y (0) = 0, y^{^{'}} (0) = 1

(d^{2} + 2 d + 3) y = e^{2} x sin (x) + 3

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