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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 10 y^{^{'}} + 16 = 0

y^{^{''}} + 2 y^{^{'}} + y = 3 x \cdot e^{- x}

y^{^{'''}} - 12 y^{^{''}} + 44 y^{^{'}} = 48

y^{^{''}} + 1 y^{^{'}} - 20 y = 1200 t^{3}

3 y^{^{'''}} + 9 y^{^{''}} + y^{^{'}} - 8 y = e^{- t}

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