Question

Determine a solução do problema de valor inicial y′−y−2^3x=0, y(0)= 1
1-Utilizando o método para EDOs Lineares de Primeira Ordem
2-Utilizando Transformada de Laplace

Answer 1

1.

$y'-y=8^x\\e^x(ye^{-x})'=e^{x\ln 8}\\\\(ye^{-x})'=e^{x(\ln 8-1)}\\\\\displaystyle\\\int_{0}^{\chi} (ye^{-x})' dx= \int_{0}^{\chi} e^{x(\ln 8-1)}dx\\\\ye^{-\chi}-1=\dfrac{1}{\ln 8-1}(8^{\chi}e^{-\chi}-1)\\\\y=\dfrac{1}{\ln 8-1}(8^{\chi}-e^{\chi})+e^{\chi}\\\\\\\boxed{y=\dfrac{8^{\chi}+e^{\chi}(\ln 8-2)}{\ln 8-1}}$

2.

$\mathcal{L}\{y'-y\}=\mathcal L \{e^{x\ln 8}\}\\ \\(s-1)\mathcal L\{y\}-1= \dfrac{1}{s-\ln 8}\\\\\\\mathcal L\{y\}=\dfrac{1}{(s-1)(s-\ln 8)}+\dfrac{1}{s-1}\\\\\\\mathcal L\{y\}=\dfrac{1}{1-\ln 8}\left(\dfrac{1}{s-1}-\dfrac{1}{s-\ln 8}\right)+\dfrac{1}{s-1}$

$y=\mathcal L^{-1}\left\{\dfrac{1}{1-\ln 8}\left(\dfrac{1}{s-1}-\dfrac{1}{s-\ln 8}\right)+\dfrac{1}{s-1}\right\}\\ \\\\y=\dfrac{2-\ln 8}{1-\ln 8}\mathcal L^{-1}\left\{\dfrac{1}{s-1}\right\}-\dfrac{1}{1-\ln 8}\mathcal L^{-1}\left\{\dfrac{1}{s-\ln 8}\right\}\\\\\\y=\dfrac{2-\ln 8}{1-\ln 8}e^{x}-\dfrac{1}{1-\ln 8}e^{\ln 8}\\ \\\\\boxed{y=\dfrac{(2-\ln 8)e^x-8^x}{1-\ln 8}}$