What is the solution for 5y^{''}+y^'=-2x ?

The solution for 5y^{''}+y^'=-2x is y=c_{1}+c_{2}e^{-x/5}-x^2+10x

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5y^{''}+y^'=-2x

5 y^{^{''}} + y^{^{'}} = - 2 x

y = c_{1} + c_{2} e^{- \frac{x}{5}} - x^{2} + 10 x

Solution steps

5 y^{''} (x) + y^{'} (x) = - 2 x

Solve linear ODE:

y = c_{1} + c_{2} e^{- \frac{x}{5}} - x^{2} + 10 x

y = c_{1} + c_{2} e^{- \frac{x}{5}} - x^{2} + 10 x

\frac{2}{s} + \frac{3}{s + 1}

\frac{d}{d x} (\frac{( 4 x ^{2} + 6 x + 4 )}{x})

\frac{( e ^{t} )}{1 + t}

x^{2} + x, at x = 1, x = 4

\frac{d}{d x} (\frac{24}{( x + 1 ) ^{5}})

What is the solution for 5y^{''}+y^'=-2x ?
The solution for 5y^{''}+y^'=-2x is y=c_{1}+c_{2}e^{-x/5}-x^2+10x