What is the solution for y^{''}+3y=2 ?

The solution for y^{''}+3y=2 is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)+2/3

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y^{''}+3y=2

y^{^{''}} + 3 y = 2

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

Solution steps

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

Solve linear ODE:

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

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

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

a^{2} \cdot y + y^{^{''}} = cos (b \cdot t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - y^{^{'}} = 1 - x^{2}

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 e^{x}

y^{^{''}} + 8 y^{^{'}} + 16 y = \frac{3 e ^{- 4 x}}{x}

What is the solution for y^{''}+3y=2 ?
The solution for y^{''}+3y=2 is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)+2/3