What is the solution for y^{''}+2y^'+5y=3e^{2x} ?

The solution for y^{''}+2y^'+5y=3e^{2x} is y=e^{-x}(c_{1}cos(2x)+c_{2}sin(2x))+3/13 e^{2x}

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y^{''}+2y^'+5y=3e^{2x}

y^{^{''}} + 2 y^{^{'}} + 5 y = 3 e^{2 x}

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{3}{13} e^{2 x}

Solution steps

y^{''} (x) + 2 y^{'} (x) + 5 y = 3 e^{2 x}

Solve linear ODE:

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{3}{13} e^{2 x}

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{3}{13} e^{2 x}

y^{^{''}} - y^{^{'}} + y = 5 t^{2}

x^{^{''}} + 2 x^{^{'}} + 5 x = 1

q^{^{''}} + 1100 q^{^{'}} + 100000 q = 16

(D^{2} + 2 D + 1) y = 4 sin (2 x)

x^{^{''}} - 16 x^{^{'}} + 64 x = 3 t e^{8 t}

What is the solution for y^{''}+2y^'+5y=3e^{2x} ?
The solution for y^{''}+2y^'+5y=3e^{2x} is y=e^{-x}(c_{1}cos(2x)+c_{2}sin(2x))+3/13 e^{2x}