What is the solution for y^{''}+5y^'+6y=4e^{-2t} ?

The solution for y^{''}+5y^'+6y=4e^{-2t} is y=c_{1}e^{-2t}+c_{2}e^{-3t}+4te^{-2t}

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y^{''}+5y^'+6y=4e^{-2t}

y^{^{''}} + 5 y^{^{'}} + 6 y = 4 e^{- 2 t}

y = c_{1} e^{- 2 t} + c_{2} e^{- 3 t} + 4 t e^{- 2 t}

Solution steps

y^{''} (t) + 5 y^{'} (t) + 6 y = 4 e^{- 2 t}

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} e^{- 3 t} + 4 t e^{- 2 t}

y = c_{1} e^{- 2 t} + c_{2} e^{- 3 t} + 4 t e^{- 2 t}

(1 + \frac{3}{y} + x) \frac{d y}{d x} + y = \frac{3}{x} - 1

arcsin (x^{2})

\int sin (4 π x) d x

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

cos (3 x), 0

What is the solution for y^{''}+5y^'+6y=4e^{-2t} ?
The solution for y^{''}+5y^'+6y=4e^{-2t} is y=c_{1}e^{-2t}+c_{2}e^{-3t}+4te^{-2t}