What is the solution for y^{''}-8y^'+16y=12x+6 ?

The solution for y^{''}-8y^'+16y=12x+6 is y=c_{1}e^{4x}+c_{2}xe^{4x}+(3x)/4+3/4

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y^{''}-8y^'+16y=12x+6

y^{^{''}} - 8 y^{^{'}} + 16 y = 12 x + 6

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

Solution steps

y^{''} (x) - 8 y^{'} (x) + 16 y = 12 x + 6

Solve linear ODE:

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

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

(d^{2} + 4) y = cos (2 x) + cos (4 x)

y^{^{'''}} - y^{^{''}} - 25 y^{^{'}} + 25 y = 9 - e^{x} + e^{5 x}

y^{^{''}} + 4 y^{^{'}} + 13 y = t e^{- t}, y (0) = 0, y^{^{'}} (0) = 4

D (T) = - 2

y^{^{''}} + 9 y = e^{t}, y (0) = 0

What is the solution for y^{''}-8y^'+16y=12x+6 ?
The solution for y^{''}-8y^'+16y=12x+6 is y=c_{1}e^{4x}+c_{2}xe^{4x}+(3x)/4+3/4