What is the solution for y^'+2y=2,6sin(3t)y(0)=3 ?

The solution for y^'+2y=2,6sin(3t)y(0)=3 is y=-e^{-2t-c_{1}}+1

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y^'+2y=2,6sin(3t)y(0)=3

y^{^{'}} + 2 y = 2, 6 sin (3 t) y (0) = 3

y = - e^{- 2 t - c_{1}} + 1

Solution steps

y^{'} (t) + 2 y = 2

Solve separable ODE:

y = - e^{- 2 t - c_{1}} + 1

y = - e^{- 2 t - c_{1}} + 1

3 x \frac{d y}{d x} + 9 x^{3} y = 3 x^{3}

x \frac{d y}{d x} + cot (y) = 0

N^{^{'}} = r - a (N - b)^{2}

y^{^{'}} = \frac{1}{2} y cos^{2} (y)

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

What is the solution for y^'+2y=2,6sin(3t)y(0)=3 ?
The solution for y^'+2y=2,6sin(3t)y(0)=3 is y=-e^{-2t-c_{1}}+1