Question

dy/dx=x/y+y/x; y(1)=-4 PLEASE HELP

Answer 1

\[\frac{ dy }{dx }=\frac{ x }{y }+\frac{ y }{x}=\frac{ x^2+y^2 }{xy }\]
put y=vx
\[\frac{ dy }{dx }=v+x \frac{ dv }{ dx }\]
\[v+x \frac{ dv }{ dx }=\frac{ x^2+v^2x^2 }{vx^2 }=\frac{ 1+v^2 }{v }\]
\[x \frac{ dv }{dx }=\frac{ 1+v^2 }{v }-v=\frac{ 1+v^2-v^2 }{v }=\frac{ 1 }{v }\]
separate the variables and integrate
\[\int\limits v~dv=\int\limits \frac{ dx }{ x },\frac{ v^2 }{ 2 }=\ln x+c\]
\[\frac{ y^2 }{x^2 }=2\ln x+C,when~x=1,y=-4,\frac{ \left( -4 \right)^2 }{ 1^2 }=2 \ln 1 +C\]
16=2*0+C,C=16
\[y^2=x^2 \ln x^2+16x^2\]