What is the solution for y^{''}+3y^'=6e^{-3t},y(0)=8,y^'(0)=-4 ?

The solution for y^{''}+3y^'=6e^{-3t},y(0)=8,y^'(0)=-4 is y=8-2/3+2/3 e^{-3t}-2te^{-3t}

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y^{''}+3y^'=6e^{-3t},y(0)=8,y^'(0)=-4

y^{^{''}} + 3 y^{^{'}} = 6 e^{- 3 t}, y (0) = 8, y^{^{'}} (0) = - 4

y = 8 - \frac{2}{3} + \frac{2}{3} e^{- 3 t} - 2 t e^{- 3 t}

Solution steps

y^{''} (t) + 3 y^{'} (t) = 6 e^{- 3 t}

Solve linear ODE:

y = 8 - \frac{2}{3} + \frac{2}{3} e^{- 3 t} - 2 t e^{- 3 t}

y = 8 - \frac{2}{3} + \frac{2}{3} e^{- 3 t} - 2 t e^{- 3 t}

\frac{d}{d x} (e^{x} x^{3} + 6 e^{x} x^{2} + 6 e^{x} x)

y^{2} d x - (1 - x) d y = 0

x \to 0 lim (3)

\frac{d}{d x} (- 3 x^{5} + 4 x^{- 3} - 5 x^{- 2} + 5)

x \to \infty lim (\frac{x}{2 a})

What is the solution for y^{''}+3y^'=6e^{-3t},y(0)=8,y^'(0)=-4 ?
The solution for y^{''}+3y^'=6e^{-3t},y(0)=8,y^'(0)=-4 is y=8-2/3+2/3 e^{-3t}-2te^{-3t}