Question

Procurar o fator integrante e resolver a equação:
(y/x)dx + (y³ - Lg x)dy = 0

Answer 1

Mdx + Ndy = 0

temos
$\boxed{\boxed{\left( \frac{y}{x}\right)dx + (y^3-ln(x))dy=0 }}$

.
$\boxed{\boxed{ \frac{\partial M}{\partial x} = \frac{y}{x^2} \;\;\;\ \left| \;\; \frac{\partial M}{\partial y} = \frac{1}{x} }}\\\\\\ \boxed{\boxed{ \frac{\partial N}{\partial x} = 0- \frac{1}{x} \;\;\;\ \left| \;\; \frac{\partial N}{\partial y} = 3y^2-0 }}\\\\.$


podemos o escrever em função de y
$g(y)= \frac{1}{M} \left( \frac{\partial M}{\partial y} -\frac{\partial N}{\partial x} \right)\\\\g(y)= \frac{x}{y} *( \frac{1}{x} + \frac{1}{x} )\\\\\boxed{\boxed{g(y)= \frac{2}{y} }}$

o fator integrante será
$I=e^{-\int g(y)dy}\\\\ I=e^{-\int \left ( \frac{2}{y} \right)dy }\\\\I=e^{-2ln(y)}\\\\ I=e^{ln(y^{-2})}\\\\\boxed{\boxed{I= \frac{1}{y^2} }}$

multiplicando a equação inicial pelo fator integrante

$\frac{1}{y^2}*( \frac{y}{x} ) dx + \frac{1}{y^2}*(y^3-ln(x))dy = 0 \\\\ \boxed{\boxed{ \left(\frac{1}{xy}\right)dx + \left( y- \frac{ln(x)}{y^2} \right)dy =0}}$

resolvendo
$F(x,y)=\int \frac{1}{xy} dx = \frac{ln(x)}{y}+ h(y)$

então
$\frac{\partial F}{\partial y}= N\\\\ - \frac{ln(x)}{y^2}+h'(y) = y- \frac{ln(x)}{y^2} \\\\\boxed{\boxed{h'(y)=y}}$

integrando o h'(y)
$h(x)=\int y\; dy = \frac{y^2}{2}+C$

logo:

$F(x,y)= \frac{ln(x)}{y}+ h(y)\\\\\boxed{\boxed{F(x,y)= \frac{ln(x)}{y}+ \frac{y^2}{2}+C_1}}$

F(x,y) é uma constante então

$C_2 = F(x,y)\\\\C_2= \frac{ln(x)}{y}+ \frac{y^2}{2}+C_1}}\\\\C_2-C_1 = \frac{ln(x)}{y}+ \frac{y^2}{2}\\\\K = \frac{ln(x)}{y}+ \frac{y^2}{2}\\\\K= \frac{2ln(x)+y^3}{2y} \\\\ 2yK - y^3 =2ln(x) \\\\ yK - \frac{y^3}{2}= ln(x) \\\\\boxed{\boxed{f(y)= e^{\frac{2yK-y^3}{2} }}}$