Question

(y ^ 2-xy) dx+x ^ 2dY = 0 y(1)=1

Answer 1

Make it exact to find a general solution.

Answer 2

resolver a equação homogenea

Answer 3

\[x ^{2}\frac{ dy }{ dx }=-y ^{2}+xy,\frac{ dy }{ dx }=\frac{ -y ^{2}+xy }{x ^{2}}\]
put y=vx
\[\frac{ dy }{ dx }=v+x \frac{ dv }{ dx }\]
\[v+x \frac{ dv }{ dx }=\frac{- v ^{2}x ^{2}+vx ^{2} }{x ^{2} }\]
\[=-v ^{2}+v\]
\[x \frac{ dv }{dx }=-v ^{2}+v-v=-v ^{2}\]
separate the variables and integrate.