What is the solution for y^{''}-12y^'+36y=12x+3 ?

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

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y^{''}-12y^'+36y=12x+3

y^{^{''}} - 12 y^{^{'}} + 36 y = 12 x + 3

y = c_{1} e^{6 x} + c_{2} x e^{6 x} + \frac{x}{3} + \frac{7}{36}

Solution steps

y^{''} (x) - 12 y^{'} (x) + 36 y = 12 x + 3

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} x e^{6 x} + \frac{x}{3} + \frac{7}{36}

y = c_{1} e^{6 x} + c_{2} x e^{6 x} + \frac{x}{3} + \frac{7}{36}

y^{^{''}} + 25 y = 3 sec (5 x)

y^{^{''}} - 4 y^{^{'}} + 20 y = 3 x^{2} - 1

4 y^{^{'''}} - y^{^{''}} = 3 x

y^{^{''}} + 16 y^{^{'}} + 64 y = (36 x^{2} - 24 x + 12) e^{- 8 x}

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{t}, y (0) = 1, y^{^{'}} (0) = - 1

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