What is the solution for y^{''}+2y^'+y=16e^{-t} ?

The solution for y^{''}+2y^'+y=16e^{-t} is y=c_{1}e^{-t}+c_{2}te^{-t}+8t^2e^{-t}

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y^{''}+2y^'+y=16e^{-t}

y^{^{''}} + 2 y^{^{'}} + y = 16 e^{- t}

y = c_{1} e^{- t} + c_{2} t e^{- t} + 8 t^{2} e^{- t}

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = 16 e^{- t}

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} t e^{- t} + 8 t^{2} e^{- t}

y = c_{1} e^{- t} + c_{2} t e^{- t} + 8 t^{2} e^{- t}

2 x + 1

\int_{- 2}^{3} \frac{37}{x ^{4}} d x

x \to 0 lim (\frac{20 sin ( 3 x )}{x})

\int \frac{1}{( x - 2 ) ^{\frac{1}{3}}} d x

h \to 0 lim \frac{( 5 ( x + h ) + 4 ) - 5 x + 4}{h}

What is the solution for y^{''}+2y^'+y=16e^{-t} ?
The solution for y^{''}+2y^'+y=16e^{-t} is y=c_{1}e^{-t}+c_{2}te^{-t}+8t^2e^{-t}