What is the solution for 2y^{''}+5y^'+3y=e^{-t} ?

The solution for 2y^{''}+5y^'+3y=e^{-t} is y=c_{1}e^{-t}+c_{2}e^{-(3t)/2}+e^{-t}t

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2y^{''}+5y^'+3y=e^{-t}

2 y^{^{''}} + 5 y^{^{'}} + 3 y = e^{- t}

y = c_{1} e^{- t} + c_{2} e^{- \frac{3 t}{2}} + e^{- t} t

Solution steps

2 y^{''} (t) + 5 y^{'} (t) + 3 y = e^{- t}

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- \frac{3 t}{2}} + e^{- t} t

y = c_{1} e^{- t} + c_{2} e^{- \frac{3 t}{2}} + e^{- t} t

a y^{^{'}} + y = 0

x \to - \infty lim (3 e^{x})

\frac{d}{d x} (x^{2} + i)

\int (1 + ln (x) + \frac{y}{x}) d x

f (x) = e^{7} + ln (6)

What is the solution for 2y^{''}+5y^'+3y=e^{-t} ?
The solution for 2y^{''}+5y^'+3y=e^{-t} is y=c_{1}e^{-t}+c_{2}e^{-(3t)/2}+e^{-t}t