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

The solution for y^{''}+4y^'+5y=e^x+5x is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+1/10 e^x+x-4/5

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

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

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{10} e^{x} + x - \frac{4}{5}

Solution steps

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

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{10} e^{x} + x - \frac{4}{5}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{10} e^{x} + x - \frac{4}{5}

\frac{1}{( s ^{2} + 4 ) \cdot s ^{2} \cdot e}

\frac{\partial}{\partial x} (sec^{2} (x))

\frac{\partial}{\partial x} (6 x^{2})

f (x) = 3 x \cdot cos (x), at x = π

\int 3 cos (n x) d x

What is the solution for y^{''}+4y^'+5y=e^x+5x ?
The solution for y^{''}+4y^'+5y=e^x+5x is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+1/10 e^x+x-4/5