What is the solution for y^{''}-3+y^'+2y=0 ?

The solution for y^{''}-3+y^'+2y=0 is y=e^{-t/2}(c_{1}cos((sqrt(7)t)/2)+c_{2}sin((sqrt(7)t)/2))+3/2

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y^{''}-3+y^'+2y=0

y^{^{''}} - 3 + y^{^{'}} + 2 y = 0

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2})) + \frac{3}{2}

Solution steps

y^{''} (t) - 3 + y^{'} (t) + 2 y = 0

Solve linear ODE:

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2})) + \frac{3}{2}

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2})) + \frac{3}{2}

y^{^{''}} - 4 y = 4 x

x^{^{''}} + 2 x^{^{'}} + 151 x = 8 cos (12 t)

y^{^{''}} + 3 y = - 48 t^{2} e^{3} t

y^{^{''''}} - 256 y = 2 e^{- 4 x} + 3 e^{3 x} + cos (4 x) + 2

y^{^{''}} + y = 3 e^{- 6 x}, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}-3+y^'+2y=0 ?
The solution for y^{''}-3+y^'+2y=0 is y=e^{-t/2}(c_{1}cos((sqrt(7)t)/2)+c_{2}sin((sqrt(7)t)/2))+3/2