What is the solution for y^{''}+4y^'=16x ?

The solution for y^{''}+4y^'=16x is y=c_{1}+c_{2}e^{-4x}+2x^2-x

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y^{''}+4y^'=16x

y^{^{''}} + 4 y^{^{'}} = 16 x

y = c_{1} + c_{2} e^{- 4 x} + 2 x^{2} - x

Solution steps

y^{''} (x) + 4 y^{'} (x) = 16 x

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 x} + 2 x^{2} - x

y = c_{1} + c_{2} e^{- 4 x} + 2 x^{2} - x

y^{^{'''}} - y^{^{''}} - 25 y^{^{'}} + 25 y = 3 - e^{x} + e^{5 x}

y^{^{''}} - 4 y^{^{'}} + 3 y = 2 x + 1

\frac{d ^{2} y}{d x ^{2}} - 7 \frac{d y}{d x} = - 3

y^{^{''}} + y^{^{'}} = 2 sin (2) t

y^{^{''}} - 6 y^{^{'}} + 25 y = 30 e^{3 x} sin (x)

What is the solution for y^{''}+4y^'=16x ?
The solution for y^{''}+4y^'=16x is y=c_{1}+c_{2}e^{-4x}+2x^2-x