What is the solution for 5y^{''}+y^'=-4x ?

The solution for 5y^{''}+y^'=-4x is y=c_{1}+c_{2}e^{-x/5}-2x^2+20x

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

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

y = c_{1} + c_{2} e^{- \frac{x}{5}} - 2 x^{2} + 20 x

Solution steps

5 y^{''} (x) + y^{'} (x) = - 4 x

Solve linear ODE:

y = c_{1} + c_{2} e^{- \frac{x}{5}} - 2 x^{2} + 20 x

y = c_{1} + c_{2} e^{- \frac{x}{5}} - 2 x^{2} + 20 x

\int \frac{5 x ^{4}}{1 + x ^{10}} d x

\int x^{2} - 6 x + 10 d x

y^{^{''}} - 5 y^{^{'}} = (8 t - 7) sin (2 t)

x \to 7 lim (x (x - 2))

\frac{d}{d x} (cos (5 x^{2} + 7 x - 1))

What is the solution for 5y^{''}+y^'=-4x ?
The solution for 5y^{''}+y^'=-4x is y=c_{1}+c_{2}e^{-x/5}-2x^2+20x