What is the solution for y^{''}-12y^'+36y=24x+6 ?

The solution for y^{''}-12y^'+36y=24x+6 is y=c_{1}e^{6x}+c_{2}xe^{6x}+(2x)/3+7/18

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y^{''}-12y^'+36y=24x+6

y^{^{''}} - 12 y^{^{'}} + 36 y = 24 x + 6

y = c_{1} e^{6 x} + c_{2} x e^{6 x} + \frac{2 x}{3} + \frac{7}{18}

Solution steps

y^{''} (x) - 12 y^{'} (x) + 36 y = 24 x + 6

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} x e^{6 x} + \frac{2 x}{3} + \frac{7}{18}

y = c_{1} e^{6 x} + c_{2} x e^{6 x} + \frac{2 x}{3} + \frac{7}{18}

y^{^{''}} - 12 y^{^{'}} + 36 y = 24 x + 5

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

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)

What is the solution for y^{''}-12y^'+36y=24x+6 ?
The solution for y^{''}-12y^'+36y=24x+6 is y=c_{1}e^{6x}+c_{2}xe^{6x}+(2x)/3+7/18