Procurar o fator integrante e resolver a equação:
y²dy + ydx - xdy = 0
Soluções para a tarefa
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Ordenemos
.....(*)
Obvio que no es una EDO exacta, pero intentemos buscar un factor integrante

Cálculo del factor integrante (FI)

Multiplicamos a la EDO en (*) por FI
![\dfrac{1}{y}\,dx+\dfrac{y^2-x}{y^2}dy=0\\ \\
f_x(x,y)=\dfrac{1}{y}\to f(x,y)=\dfrac{x}{y}+\phi(y)\\ \\
f_y(x,y)=\dfrac{d}{dy}\left[\dfrac{x}{y}+\phi(y)\right]\\ \\
\dfrac{y^2-x}{y^2}=-\dfrac{x}{y^2}+\phi'(y)\\ \\
\phi'(y) = 1\\ \\
\phi(y)=y\\ \\\text{Entonces:}\\ \\
f(x,y)=\dfrac{x}{y}+y \dfrac{1}{y}\,dx+\dfrac{y^2-x}{y^2}dy=0\\ \\
f_x(x,y)=\dfrac{1}{y}\to f(x,y)=\dfrac{x}{y}+\phi(y)\\ \\
f_y(x,y)=\dfrac{d}{dy}\left[\dfrac{x}{y}+\phi(y)\right]\\ \\
\dfrac{y^2-x}{y^2}=-\dfrac{x}{y^2}+\phi'(y)\\ \\
\phi'(y) = 1\\ \\
\phi(y)=y\\ \\\text{Entonces:}\\ \\
f(x,y)=\dfrac{x}{y}+y](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7By%7D%5C%2Cdx%2B%5Cdfrac%7By%5E2-x%7D%7By%5E2%7Ddy%3D0%5C%5C+%5C%5C%0Af_x%28x%2Cy%29%3D%5Cdfrac%7B1%7D%7By%7D%5Cto+f%28x%2Cy%29%3D%5Cdfrac%7Bx%7D%7By%7D%2B%5Cphi%28y%29%5C%5C+%5C%5C%0Af_y%28x%2Cy%29%3D%5Cdfrac%7Bd%7D%7Bdy%7D%5Cleft%5B%5Cdfrac%7Bx%7D%7By%7D%2B%5Cphi%28y%29%5Cright%5D%5C%5C+%5C%5C%0A%5Cdfrac%7By%5E2-x%7D%7By%5E2%7D%3D-%5Cdfrac%7Bx%7D%7By%5E2%7D%2B%5Cphi%27%28y%29%5C%5C+%5C%5C%0A%5Cphi%27%28y%29+%3D+1%5C%5C+%5C%5C%0A%5Cphi%28y%29%3Dy%5C%5C+%5C%5C%5Ctext%7BEntonces%3A%7D%5C%5C+%5C%5C%0Af%28x%2Cy%29%3D%5Cdfrac%7Bx%7D%7By%7D%2By)
Por lo tanto la solución es

Obvio que no es una EDO exacta, pero intentemos buscar un factor integrante
Cálculo del factor integrante (FI)
Multiplicamos a la EDO en (*) por FI
Por lo tanto la solución es
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