What is the solution for y^{''}-2y^'-8y=te^{-4t} ?

The solution for y^{''}-2y^'-8y=te^{-4t} is y=c_{1}e^{4t}+c_{2}e^{-2t}+(e^{-4t}t)/(16)+5/128 e^{-4t}

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y^{''}-2y^'-8y=te^{-4t}

y^{^{''}} - 2 y^{^{'}} - 8 y = t e^{- 4 t}

y = c_{1} e^{4 t} + c_{2} e^{- 2 t} + \frac{e ^{- 4 t} t}{16} + \frac{5}{128} e^{- 4 t}

Solution steps

y^{''} (t) - 2 y^{'} (t) - 8 y = t e^{- 4 t}

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{- 2 t} + \frac{e ^{- 4 t} t}{16} + \frac{5}{128} e^{- 4 t}

y = c_{1} e^{4 t} + c_{2} e^{- 2 t} + \frac{e ^{- 4 t} t}{16} + \frac{5}{128} e^{- 4 t}

y^{^{''}} + y^{^{'}} + y = 2 sin (3 x)

2 \frac{d ^{2} x}{d t ^{2}} + 18 x + 5 = 0

y^{^{''}} + 9 y = t

y^{^{''}} + 2 y^{^{'}} + y = 3 x^{4} e^{- x}

9 y^{^{''}} - 6 y^{^{'}} + y = 9 - x^{2}

What is the solution for y^{''}-2y^'-8y=te^{-4t} ?
The solution for y^{''}-2y^'-8y=te^{-4t} is y=c_{1}e^{4t}+c_{2}e^{-2t}+(e^{-4t}t)/(16)+5/128 e^{-4t}