Question

Resolve a Questão 33 Pra Min

Answer 1

Questão 33:

$\dfrac{dy}{dx}=\dfrac{2}{x}-\dfrac{1}{x^{2}}\,,~~~\text{sendo que }~y_{(x\,=\,1)}=-1.$


Integrando os dois lados em $x\,,$ temos

$\displaystyle \int\frac{dy}{dx}\,dx=\int\left(\frac{2}{x}-\frac{1}{x^{2}} \right )dx\\\\\\ y=\int(2x^{-1}-x^{-2})\,dx\\\\\\ y=2\,\mathrm{\ell n\,}|x|-\dfrac{x^{-2+1}}{-2+1}+C\\\\\\ y=2\,\mathrm{\ell n\,}|x|-\dfrac{x^{-1}}{-1}+C\\\\\\ \boxed{\begin{array}{c} y=2\,\mathrm{\ell n\,}|x|+\dfrac{1}{x}+C \end{array}}~~~~~~\mathbf{(i)}$


sendo $C$ uma constante a determinar.

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Temos que $y=-1\,,$ quando $x=1.$ Portanto,

$-1=2\,\mathrm{\ell n\,}|1|+\dfrac{1}{1}+C\\\\\\ -1=2\cdot 0+1+C\\\\ C=-1-1\\\\ \boxed{\begin{array}{c} C=-2 \end{array}}$
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Substituindo em $\mathbf{(i)},$ o valor de $C$ encontrado, obtemos

$\boxed{\begin{array}{c} y=2\,\mathrm{\ell n\,}|x|+\dfrac{1}{x}-2 \end{array}}$