What is the solution for y^{''}+4y^'+5y=35e^{-4x} ?

The solution for y^{''}+4y^'+5y=35e^{-4x} is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+7e^{-4x}

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y^{''}+4y^'+5y=35e^{-4x}

y^{^{''}} + 4 y^{^{'}} + 5 y = 35 e^{- 4 x}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + 7 e^{- 4 x}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 5 y = 35 e^{- 4 x}

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + 7 e^{- 4 x}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + 7 e^{- 4 x}

\int 2 x^{2} - 3 d x

y = \frac{cos ( 4 x )}{x ^{2}}

\int - 8 x^{2} e^{x} d x

x^{\frac{3}{2}}

x \to 3 - lim (\frac{9}{( x - 3 ) ^{5}})

What is the solution for y^{''}+4y^'+5y=35e^{-4x} ?
The solution for y^{''}+4y^'+5y=35e^{-4x} is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+7e^{-4x}