What is the solution for (4D^2-4D+1)y=100 ?

The solution for (4D^2-4D+1)y=100 is y=c_{1}e^{x/2}+c_{2}e^{x/2}x+100

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(4D^2-4D+1)y=100

(4 D^{2} - 4 D + 1) y = 100

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + 100

Solution steps

(4 D^{2} - 4 D + 1) y = 100

Solve linear ODE:

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + 100

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + 100

y^{^{''}} + 2 y^{^{'}} + 4 y = x e^{x}

y^{^{''}} + 5 y^{^{'}} + 6 y = 10 (1 - x) e^{- 2 x}

y^{^{''}} + 25 y = 3, y (0) = 1, y (0) = 5

y^{^{''}} + 6 y^{^{'}} + 8 y = 2 t - 3 cos (3 t)

x^{^{''}} + 10 x^{^{'}} + 25 x = 5 sin (3 t)

What is the solution for (4D^2-4D+1)y=100 ?
The solution for (4D^2-4D+1)y=100 is y=c_{1}e^{x/2}+c_{2}e^{x/2}x+100