What is the solution for y^{''}-12y^'+36y=24x+2 ?

The solution for y^{''}-12y^'+36y=24x+2 is y=c_{1}e^{6x}+c_{2}xe^{6x}+(2x)/3+5/18

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y^{''}-12y^'+36y=24x+2

y^{^{''}} - 12 y^{^{'}} + 36 y = 24 x + 2

y = c_{1} e^{6 x} + c_{2} x e^{6 x} + \frac{2 x}{3} + \frac{5}{18}

Solution steps

y^{''} (x) - 12 y^{'} (x) + 36 y = 24 x + 2

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} x e^{6 x} + \frac{2 x}{3} + \frac{5}{18}

y = c_{1} e^{6 x} + c_{2} x e^{6 x} + \frac{2 x}{3} + \frac{5}{18}

y^{^{''}} + y = x^{2} + 1, y (0) = 5, y (1) = 0

y^{^{''}} + 25 y = 10 sin (5 t)

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

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

x^{^{''}} + 100 x^{^{'}} + 2500 x = 30

What is the solution for y^{''}-12y^'+36y=24x+2 ?
The solution for y^{''}-12y^'+36y=24x+2 is y=c_{1}e^{6x}+c_{2}xe^{6x}+(2x)/3+5/18