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

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

Answered by benjo_mathlover last updated on 23/Oct/20

$$(D+I)(D-2I)(D+2I)\left(y\right)=e^{4x}$$$$let(D-2I)(D+2I)\left(y\right)=z$$$$\Rightarrow (D+I)\left(z\right)=e^{4x}$$$$\Rightarrow z^{\prime }+z=e^{4x};e^x(z^{\prime }+z)=e^{5x}$$$$\Rightarrow {(e^x.z)}^{\prime }=e^{5x};e^x.z=\frac{1}{5}{e}^{5x}+A$$$$\Rightarrow z=\frac{1}{5}{e}^{4x}+{Ae}^{-x}$$$$\iff (D-2I)(D+2I)\left(y\right)=\frac{1}{5}{e}^{4x}+{Ae}^{-x}$$$$let(D+2I)\left(y\right)=q$$$$\Rightarrow (D-2I)\left(q\right)=\frac{1}{5}{e}^{4x}+{Ae}^{-x}$$$$\Rightarrow q^{\prime }-2q=\frac{1}{5}{e}^{4x}+{Ae}^{-x}$$$$\Rightarrow e^{-2x}(q^{\prime }-2q)=\frac{1}{5}{e}^{2x}+{Ae}^{-3x}$$$$\Rightarrow {(e^{-2x}.q)}^{\prime }=\frac{1}{5}{e}^{2x}+{Ae}^{-3x}$$$$\Rightarrow e^{-2x}.q=\int (\frac{1}{5}{e}^{2x}+{Ae}^{-3x})dx$$$$\Rightarrow q=\frac{1}{10}{e}^{4x}-\frac{1}{3}{Ae}^{-x}+{Be}^{2x}$$$$\Rightarrow (D+2I)\left(y\right)=\frac{1}{10}{e}^{4x}-\frac{1}{3}{Ae}^{-x}+{Be}^{2x}$$$$\Rightarrow e^{2x}(y^{\prime }+2y)=\frac{1}{10}{e}^{6x}-\frac{1}{3}{Ae}^x+{Be}^{4x}$$$$\Rightarrow {(e^{2x}.y)}^{\prime }=\frac{1}{10}{e}^{6x}-\frac{1}{3}{Ae}^x+{Be}^{4x}$$$$\Rightarrow e^{2x}.y=\int (\frac{1}{10}{e}^{6x}-\frac{1}{3}{Ae}^x+{Be}^{4x})dx$$$$\Rightarrow y=\frac{1}{60}{e}^{4x}-\frac{1}{3}{Ae}^{-x}+\frac{1}{4}{Be}^{2x}+{Ce}^{-2x}$$$$[-\frac{1}{3}A={\lambda }_1,\frac{1}{4}B={\lambda }_2,C={\lambda }_3]$$$$\because y=\frac{1}{60}{e}^{4x}+{\lambda }_1{e}^{-x}+{\lambda }_2{e}^{2x}+{\lambda }_3{e}^{-2x}$$

Commented by bemath last updated on 23/Oct/20

$$nice...$$

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