D-3-D-2-4D-4-y-e-4x-

Question Number 119235 by bemath last updated on 23/Oct/20

(D^{3} + D^{2} - 4 D - 4) y = e^{4 x}

Answered by benjo_mathlover last updated on 23/Oct/20

(D + I) (D - 2 I) (D + 2 I) (y) = e^{4 x}

l e t (D - 2 I) (D + 2 I) (y) = z

\Rightarrow (D + I) (z) = e^{4 x}

\Rightarrow z^{'} + z = e^{4 x}; e^{x} (z^{'} + z) = e^{5 x}

\Rightarrow {(e^{x} . z)}^{'} = e^{5 x}; e^{x} . z = \frac{1}{5} e^{5 x} + A

\Rightarrow z = \frac{1}{5} e^{4 x} + {A e}^{- x}

\Leftrightarrow (D - 2 I) (D + 2 I) (y) = \frac{1}{5} e^{4 x} + {A e}^{- x}

l e t (D + 2 I) (y) = q

\Rightarrow (D - 2 I) (q) = \frac{1}{5} e^{4 x} + {A e}^{- x}

\Rightarrow q^{'} - 2 q = \frac{1}{5} e^{4 x} + {A e}^{- x}

\Rightarrow e^{- 2 x} (q^{'} - 2 q) = \frac{1}{5} e^{2 x} + {A e}^{- 3 x}

\Rightarrow {(e^{- 2 x} . q)}^{'} = \frac{1}{5} e^{2 x} + {A e}^{- 3 x}

\Rightarrow e^{- 2 x} . q = \int (\frac{1}{5} e^{2 x} + {A e}^{- 3 x}) d x

\Rightarrow q = \frac{1}{10} e^{4 x} - \frac{1}{3} {A e}^{- x} + {B e}^{2 x}

\Rightarrow (D + 2 I) (y) = \frac{1}{10} e^{4 x} - \frac{1}{3} {A e}^{- x} + {B e}^{2 x}

\Rightarrow e^{2 x} (y^{'} + 2 y) = \frac{1}{10} e^{6 x} - \frac{1}{3} {A e}^{x} + {B e}^{4 x}

\Rightarrow {(e^{2 x} . y)}^{'} = \frac{1}{10} e^{6 x} - \frac{1}{3} {A e}^{x} + {B e}^{4 x}

\Rightarrow e^{2 x} . y = \int (\frac{1}{10} e^{6 x} - \frac{1}{3} {A e}^{x} + {B e}^{4 x}) d x

\Rightarrow y = \frac{1}{60} e^{4 x} - \frac{1}{3} {A e}^{- x} + \frac{1}{4} {B e}^{2 x} + {C e}^{- 2 x}

[- \frac{1}{3} A = λ_{1}, \frac{1}{4} B = λ_{2}, C = λ_{3}]

∵ y = \frac{1}{60} e^{4 x} + λ_{1} e^{- x} + λ_{2} e^{2 x} + λ_{3} e^{- 2 x}

Commented by bemath last updated on 23/Oct/20

n i c e \dots

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