What is the solution for y^{''}+4y^'+5y=e^{2x}+1 ?

The solution for y^{''}+4y^'+5y=e^{2x}+1 is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+1/17 e^{2x}+1/5

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y^{''}+4y^'+5y=e^{2x}+1

y^{^{''}} + 4 y^{^{'}} + 5 y = e^{2 x} + 1

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{17} e^{2 x} + \frac{1}{5}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 5 y = e^{2 x} + 1

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{17} e^{2 x} + \frac{1}{5}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{17} e^{2 x} + \frac{1}{5}

4 y^{^{''}} - y = e^{\frac{x}{2}} x

\frac{d ^{2} y}{d x ^{2}} + y = 3 sin (2 x) + x cos (2 x)

y^{^{''}} - 6 y^{^{'}} + 8 y = e^{4 x}

y^{^{''}} + 4 y^{^{'}} = 3 sin (3 t)

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

What is the solution for y^{''}+4y^'+5y=e^{2x}+1 ?
The solution for y^{''}+4y^'+5y=e^{2x}+1 is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+1/17 e^{2x}+1/5