What is the solution for y^{''}+2y^'+10y=25x^2+3 ?

The solution for y^{''}+2y^'+10y=25x^2+3 is y=e^{-x}(c_{1}cos(3x)+c_{2}sin(3x))+(5x^2)/2-x

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y^{''}+2y^'+10y=25x^2+3

y^{^{''}} + 2 y^{^{'}} + 10 y = 25 x^{2} + 3

y = e^{- x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{5 x ^{2}}{2} - x

Solution steps

y^{''} (x) + 2 y^{'} (x) + 10 y = 25 x^{2} + 3

Solve linear ODE:

y = e^{- x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{5 x ^{2}}{2} - x

y = e^{- x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{5 x ^{2}}{2} - x

(d^{3} - d^{2} - 4 d + 4) y = e^{3} x

(d + 1)^{3} y = x^{2} e^{- x}

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

y^{^{'''}} + 3 y^{^{''}} + 3 y^{^{'}} + y = 30 e^{- x}

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3} x

What is the solution for y^{''}+2y^'+10y=25x^2+3 ?
The solution for y^{''}+2y^'+10y=25x^2+3 is y=e^{-x}(c_{1}cos(3x)+c_{2}sin(3x))+(5x^2)/2-x