Question

Resolva o problema de valor inicial:

Answer 1

$\displaystyle \sf \large \left \{ {{\displaystyle y'=\frac{1}{x^2}-\frac{1}{x^3}} \atop \atop {\displaystyle y(1)=\frac{3}{2}}} \right.$

Temos :

$\displaystyle \sf y'=\frac{1}{x^2}-\frac{1}{x^3} \\\\ \underline{integrando \ dos \ dois \ lados}: \\\\\\ y = \int x^{-2}dx -\int x^{-3}dx\\\\ y = \frac{x^{-2+1}}{-2+1}-\frac{x^{-3+1}}{-3+1}+C\\\\\\ y = \frac{-1}{x} + \frac{1}{2x^2}+ C \\\\\\ \underline{usando \ y(1) = \frac{3}{2}}: \\\\\\ \frac{3}{2}=\frac{-1}{1}+\frac{1}{2.1^2}+C \\\\\\ C = \frac{3}{2}-\frac{1}{2}+1 \\\\\\ C =2$

Portanto :

$\huge\boxed{\sf \ y = -\frac{1}{x}+\frac{1}{2x^2}+2\ }\checkmark$