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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 30, y (0) = 0, y^{^{'}} (0) = 0

q^{^{''}} + q^{^{'}} + \frac{1}{0.05} = 80 t

y^{^{''}} - y = - 11 t + 1

y^{^{''}} - 3 y^{^{'}} - 46 y = 2 sin (t)

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