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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y = 28

y^{^{'''}} + 196 y^{^{'}} = tan (14 x)

y^{^{''}} + 4 y^{^{'}} + 16 y = 4 sin (6 t)

y^{^{''}} + 16 y = 16, y (0) = 1, y^{^{'}} (0) = 4

y^{^{''}} + 5 y^{^{'}} + 4 y = e^{- x} + 2

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