What is the solution for y^{''}-2y^'=-4y ?

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

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y^{''}-2y^'=-4y

y^{^{''}} - 2 y^{^{'}} = - 4 y

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

Solution steps

y^{''} (t) - 2 y^{'} (t) = - 4 y

Solve linear ODE:

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

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

7 y^{^{''}} - 8 y^{^{'}} + y = 0

60 x^{^{''}} + 240 x = 0, x (0) = 0, x^{^{'}} (0) = - 8

4 y^{^{''}} + 9 y^{^{'}} + 9 y = 0

y^{^{''}} - 10 y^{^{'}} - 144 y = 0, y (0) = 9, y^{^{'}} (0) = 58

y^{^{''}} + a \cdot y^{^{'}} - b \cdot y = 0

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