What is the solution for 5y^{''}+y^'=-8x ?

The solution for 5y^{''}+y^'=-8x is y=c_{1}+c_{2}e^{-x/5}-4x^2+40x

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5y^{''}+y^'=-8x

5 y^{^{''}} + y^{^{'}} = - 8 x

y = c_{1} + c_{2} e^{- \frac{x}{5}} - 4 x^{2} + 40 x

Solution steps

5 y^{''} (x) + y^{'} (x) = - 8 x

Solve linear ODE:

y = c_{1} + c_{2} e^{- \frac{x}{5}} - 4 x^{2} + 40 x

y = c_{1} + c_{2} e^{- \frac{x}{5}} - 4 x^{2} + 40 x

\int \frac{1}{7 - 6 x - x ^{2}} d x

7 x^{2}

\frac{\partial}{\partial x} (x e^{y x^{2}} sin (x))

\int_{0}^{1} 1

\int \frac{3}{2} d x

What is the solution for 5y^{''}+y^'=-8x ?
The solution for 5y^{''}+y^'=-8x is y=c_{1}+c_{2}e^{-x/5}-4x^2+40x