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

The solution for y^{''}-4y^'+4y=6x+4 is y=c_{1}e^{2x}+c_{2}xe^{2x}+(3x)/2+5/2

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

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

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{3 x}{2} + \frac{5}{2}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{3 x}{2} + \frac{5}{2}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{3 x}{2} + \frac{5}{2}

y^{^{''}} + 4 y^{^{'}} + 3 y = 2 e^{- 2 t}

y^{^{''}} - 10 y^{^{'}} + 16 y = 17 sin (2 x)

y^{^{''}} + 25 y = 12 cos (5 x) + 18 sin (5 x)

y^{^{''}} + 8 y^{^{'}} + 7 y = 16 x^{2}

y^{^{''}} - 6 y^{^{'}} + 9 y = x e^{3 x}

What is the solution for y^{''}-4y^'+4y=6x+4 ?
The solution for y^{''}-4y^'+4y=6x+4 is y=c_{1}e^{2x}+c_{2}xe^{2x}+(3x)/2+5/2