What is the solution for y^{''}+2y^'+y=e^{-t}+9e^{5t} ?

The solution for y^{''}+2y^'+y=e^{-t}+9e^{5t} is y=c_{1}e^{-t}+c_{2}te^{-t}+(e^{-t}t^2)/2+1/4 e^{5t}

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y^{''}+2y^'+y=e^{-t}+9e^{5t}

y^{^{''}} + 2 y^{^{'}} + y = e^{- t} + 9 e^{5 t}

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{e ^{- t} t ^{2}}{2} + \frac{1}{4} e^{5 t}

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = e^{- t} + 9 e^{5 t}

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{e ^{- t} t ^{2}}{2} + \frac{1}{4} e^{5 t}

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{e ^{- t} t ^{2}}{2} + \frac{1}{4} e^{5 t}

n \to \infty lim (\frac{i}{n})

(1 + z)^{\frac{1}{4}}

\frac{d}{d x} (5 x)

\frac{\partial}{\partial x} (y^{2} - cos (x) y)

\frac{d}{d x} (540 (1 - e^{- 0.1 x}))

What is the solution for y^{''}+2y^'+y=e^{-t}+9e^{5t} ?
The solution for y^{''}+2y^'+y=e^{-t}+9e^{5t} is y=c_{1}e^{-t}+c_{2}te^{-t}+(e^{-t}t^2)/2+1/4 e^{5t}