What is the solution for y^{''}-4y^'+4y=6x^2 ?

The solution for y^{''}-4y^'+4y=6x^2 is y=c_{1}e^{2x}+c_{2}xe^{2x}+(3x^2)/2+3x+9/4

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y^{''}-4y^'+4y=6x^2

y^{^{''}} - 4 y^{^{'}} + 4 y = 6 x^{2}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{3 x ^{2}}{2} + 3 x + \frac{9}{4}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = 6 x^{2}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{3 x ^{2}}{2} + 3 x + \frac{9}{4}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{3 x ^{2}}{2} + 3 x + \frac{9}{4}

y^{^{''}} + y^{^{'}} = 7 sin (x)

y^{^{''}} + 2 y^{^{'}} - 1 = 0

y^{^{''}} + 4 y^{^{'}} + 4 y = x^{2} \cdot e^{(- 2 x)} + 1

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

y^{^{''}} - y = sin (t), < y (0) = 0, y^{^{'}} (0) = 1

What is the solution for y^{''}-4y^'+4y=6x^2 ?
The solution for y^{''}-4y^'+4y=6x^2 is y=c_{1}e^{2x}+c_{2}xe^{2x}+(3x^2)/2+3x+9/4