What is the solution for y^{''}-12y^'+36y=7+2x ?

The solution for y^{''}-12y^'+36y=7+2x is y=c_{1}e^{6x}+c_{2}xe^{6x}+x/(18)+23/108

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y^{''}-12y^'+36y=7+2x

y^{^{''}} - 12 y^{^{'}} + 36 y = 7 + 2 x

y = c_{1} e^{6 x} + c_{2} x e^{6 x} + \frac{x}{18} + \frac{23}{108}

Solution steps

y^{''} (x) - 12 y^{'} (x) + 36 y = 7 + 2 x

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} x e^{6 x} + \frac{x}{18} + \frac{23}{108}

y = c_{1} e^{6 x} + c_{2} x e^{6 x} + \frac{x}{18} + \frac{23}{108}

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

y^{^{''}} - 3 y = cos (3 x)

y^{^{''}} - 4 y = 4 t - 8 e^{- 2 t}

y^{^{''}} + y^{^{'}} = cos (3 t), y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 8 y^{^{'}} + 16 y = \frac{3 e ^{- 4 x}}{x ^{4}}

What is the solution for y^{''}-12y^'+36y=7+2x ?
The solution for y^{''}-12y^'+36y=7+2x is y=c_{1}e^{6x}+c_{2}xe^{6x}+x/(18)+23/108