What is the solution for y^{''}-2y^'=12x+10 ?

The solution for y^{''}-2y^'=12x+10 is y=c_{1}e^{2x}+c_{2}-3x^2-8x

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y^{''}-2y^'=12x+10

y^{^{''}} - 2 y^{^{'}} = 12 x + 10

y = c_{1} e^{2 x} + c_{2} - 3 x^{2} - 8 x

Solution steps

y^{''} (x) - 2 y^{'} (x) = 12 x + 10

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} - 3 x^{2} - 8 x

y = c_{1} e^{2 x} + c_{2} - 3 x^{2} - 8 x

y^{^{''}} + 2 y^{^{'}} + 17 y = 20 e^{t}, y (0) = 2, y^{^{''}} (0) = - 4

y^{^{''}} + 7 y^{^{'}} + 12 y = 2 e^{- t}

y^{^{''}} + 16 y = sin (8 x)

2 x^{^{''}} + 3 x^{^{'}} + 5 x = 2 t

y^{^{''}} - 6 y^{^{'}} + 8 y = 8 e^{4 x}

What is the solution for y^{''}-2y^'=12x+10 ?
The solution for y^{''}-2y^'=12x+10 is y=c_{1}e^{2x}+c_{2}-3x^2-8x