Question

Calcular a equação diferencial homogênea utilizando a substituição
não adianta responder só pra ganhar pts pq vou denunciar e mandar pra um moderador

Answer 1

$\large\boxed{\begin{array}{l}\rm\dfrac{dy}{dx}=\dfrac{x-y}{x}\\\\\rm\dfrac{dy}{dx}=1-\dfrac{y}{x}\\\\\rm\dfrac{dy}{dx}+\dfrac{1}{x}\cdot y=1\longrightarrow equac_{\!\!,}\tilde ao\,diferencial\,linear\\\\\underline{\sf C\acute alculo\,do\,fator\,integrante:}\\\rm \mu(x)=e^{\displaystyle\int\dfrac{1}{x}\,dx}=e^{\ln x}=x\\\sf multiplicando\,a\,equac_{\!\!,}\tilde ao\,diferencial\,por\,\mu(x)\,temos:\\\rm\mu(x)\cdot\dfrac{dy}{dx}+\mu(x)\cdot\dfrac{1}{x}y=\mu(x)\cdot 1\end{array}}$

$\large\boxed{\begin{array}{l}\rm x\dfrac{dy}{dx}+\diagup\!\!\! x\cdot\dfrac{1}{\diagup\!\!\!x}y=x\\\rm\dfrac{d}{dx}(y\cdot x)=x\\\\\rm d(y\cdot x)=x\,dx\\\displaystyle\rm\int d(y\cdot x)=\int x\,dx\\\rm y\cdot x=\dfrac{1}{2}x^2+k\\\sf isolando\,y\,temos:\\\rm y(x)=\dfrac{1}{2}\diagup\!\!\!\!x^2\cdot\dfrac{1}{\diagup\!\!\!\!x}+\dfrac{k}{x}\\\\\rm y(x)=\dfrac{x}{2}+\dfrac{k}{x}\end{array}}$