What is the solution for y^{''}-12y^'+36y=12x+5 ?

The solution for y^{''}-12y^'+36y=12x+5 is y=c_{1}e^{6x}+c_{2}xe^{6x}+x/3+1/4

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

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

y = c_{1} e^{6 x} + c_{2} x e^{6 x} + \frac{x}{3} + \frac{1}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} x e^{6 x} + \frac{x}{3} + \frac{1}{4}

y = c_{1} e^{6 x} + c_{2} x e^{6 x} + \frac{x}{3} + \frac{1}{4}

y^{^{''}} + y = sin (x) + cos (x)

y^{^{''}} + 4 y = t^{2} + 2 e^{t}, y (0) = 0, y^{^{'}} (0) = 3

y^{^{''}} + 2 y^{^{'}} + 2 y = 2 cos (2 t)

y^{^{''}} + 2 y^{^{'}} + 5 y = 2 sin (t)

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

What is the solution for y^{''}-12y^'+36y=12x+5 ?
The solution for y^{''}-12y^'+36y=12x+5 is y=c_{1}e^{6x}+c_{2}xe^{6x}+x/3+1/4