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))

y^{^{''}} + y^{^{'}} + 5 y = 0, y (0) = 1.2, y^{^{'}} (0) = 0

(x - 1) y^{^{''}} - x y + y = 0

3 \frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + y = 0

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

D^{2} (D^{2} - 4 D + 5)^{3} 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))