What is the solution for 2y^{''}+y^'=4x ?

The solution for 2y^{''}+y^'=4x is y=c_{1}+c_{2}e^{-x/2}+2x^2-8x

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

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

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

Solution steps

2 y^{''} (x) + y^{'} (x) = 4 x

Solve linear ODE:

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

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

What is the solution for 2y^{''}+y^'=4x ?
The solution for 2y^{''}+y^'=4x is y=c_{1}+c_{2}e^{-x/2}+2x^2-8x