What is the solution for y^{''}+2y^'+y=4e-t ?

The solution for y^{''}+2y^'+y=4e-t is y=c_{1}e^{-t}+c_{2}te^{-t}-t+4e+2

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y^{''}+2y^'+y=4e-t

y^{^{''}} + 2 y^{^{'}} + y = 4 e - t

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = 4 e - t

Solve linear ODE:

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

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

1 y^{2} - 7 y^{4} (y + 9 y^{3})

x \to 36 lim (\frac{x ^{2}}{9} - \frac{4}{x})

\int \frac{7}{4 + 7 x} d x

\frac{\partial}{\partial x} (3 x^{2} y + x sin (2 y))

y^{^{'}} = \frac{3 y ^{2} - x ^{2}}{2 x y}

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