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

The solution for y^{''}+2y^'+10y=10x+2 is y=e^{-x}(c_{1}cos(3x)+c_{2}sin(3x))+x

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

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

y = e^{- x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + x

Solution steps

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

Solve linear ODE:

y = e^{- x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + x

y = e^{- x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + x

y^{^{'''}} + 3 y^{^{''}} - 7 y^{^{'}} + 2 y = 3 x + 10 sin (x)

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

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

y^{^{'''}} - 2 y^{^{''}} + y^{^{'}} = 5 + 4 x e^{x}

1.2 x^{^{''}} + 1600 x^{^{'}} + 100000 = 18

What is the solution for y^{''}+2y^'+10y=10x+2 ?
The solution for y^{''}+2y^'+10y=10x+2 is y=e^{-x}(c_{1}cos(3x)+c_{2}sin(3x))+x