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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = 5, y^{^{'}} (0) = 13

y^{^{'''}} + 3 y - 10 y^{^{'}} = 0

(D^{3} + 6 D^{2} + 11 D + 6) y = 0

y^{^{''}} + \frac{48}{2} y^{^{'}} = 0

\frac{d ^{2} y}{d t ^{2}} - 8 \frac{d y}{d t} + 16 y = 0

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