What is the solution for (D^2-2D+2)y=0 ?

The solution for (D^2-2D+2)y=0 is y=e^x(c_{1}cos(x)+c_{2}sin(x))

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(D^2-2D+2)y=0

(D^{2} - 2 D + 2) y = 0

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

Solution steps

(D^{2} - 2 D + 2) y = 0

Solve linear ODE:

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

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

4 \frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} = - y

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

\frac{d ^{2} y}{d x ^{2}} - 8 \frac{d y}{d x} + 15 y = 0

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

y^{^{''}} + 17 y^{^{'}} + 72 y = 0

What is the solution for (D^2-2D+2)y=0 ?
The solution for (D^2-2D+2)y=0 is y=e^x(c_{1}cos(x)+c_{2}sin(x))