What is the solution for y^{''}+2y^'+2y=5e^{6x} ?

The solution for y^{''}+2y^'+2y=5e^{6x} is y=e^{-x}(c_{1}cos(x)+c_{2}sin(x))+1/10 e^{6x}

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y^{''}+2y^'+2y=5e^{6x}

y^{^{''}} + 2 y^{^{'}} + 2 y = 5 e^{6 x}

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{10} e^{6 x}

Solution steps

y^{''} (x) + 2 y^{'} (x) + 2 y = 5 e^{6 x}

Solve linear ODE:

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{10} e^{6 x}

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{10} e^{6 x}

\frac{d ^{2} x}{d t ^{2}} + x k + a = 0

y^{^{''}} - 2 y^{^{'}} + y = x^{\frac{1}{2}} e^{x}

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

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

y^{^{''}} - 6 y^{^{'}} + 9 y = t^{2} e^{3 t}

What is the solution for y^{''}+2y^'+2y=5e^{6x} ?
The solution for y^{''}+2y^'+2y=5e^{6x} is y=e^{-x}(c_{1}cos(x)+c_{2}sin(x))+1/10 e^{6x}