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

The solution for y^{''}+2y^'+5y=x is y=e^{-x}(c_{1}cos(2x)+c_{2}sin(2x))+x/5-2/25

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

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

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{x}{5} - \frac{2}{25}

Solution steps

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

Solve linear ODE:

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{x}{5} - \frac{2}{25}

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{x}{5} - \frac{2}{25}

\int \frac{5}{( x ^{2} - 1 ) ^{2}} d x

\frac{6}{x ^{4}}

\int 2 t^{3} - 6 t + 7 d t

\frac{d}{d x} (\frac{14 x + 6 x ^{2}}{( 7 + 6 x ) ^{2}})

\frac{\partial}{\partial x} (\frac{- x - 7 y}{y ^{2} + 6 x ^{2}})

What is the solution for y^{''}+2y^'+5y=x ?
The solution for y^{''}+2y^'+5y=x is y=e^{-x}(c_{1}cos(2x)+c_{2}sin(2x))+x/5-2/25