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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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