What is the solution for (D^3-4D^2+4D)y=0 ?

The solution for (D^3-4D^2+4D)y=0 is y=c_{1}+c_{2}e^{2x}+c_{3}xe^{2x}

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(D^3-4D^2+4D)y=0

(D^{3} - 4 D^{2} + 4 D) y = 0

y = c_{1} + c_{2} e^{2 x} + c_{3} x e^{2 x}

Solution steps

(D^{3} - 4 D^{2} + 4 D) y = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{2 x} + c_{3} x e^{2 x}

y = c_{1} + c_{2} e^{2 x} + c_{3} x e^{2 x}

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = - 3, y^{^{'}} (0) = 7

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

2 y^{^{''}} + 2 y^{^{'}} + 5 y = 0

16 \frac{d ^{4} y}{d x ^{4}} - y = 0

x^{^{''}} + 8 x^{^{'}} + 4 x = 0, x (0) = 0, x^{^{'}} (0) = 1

What is the solution for (D^3-4D^2+4D)y=0 ?
The solution for (D^3-4D^2+4D)y=0 is y=c_{1}+c_{2}e^{2x}+c_{3}xe^{2x}