<s>[INST] Substitution to solve an initial value problem By using the substitution $y(x) = v(x)x$, how can I solve the initial value problem
$$
\frac{dy}{dx} = \frac{x^2+y^2}{xy - x^2},\quad y(1)=1
$$
And also keep my answer in the form $g(x,y)= 4e^{-1} xe^\frac{y}{x}$
 [/INST] $$\frac { dy }{ dx } =\frac { { x }^{ 2 }+{ y }^{ 2 } }{ xy-{ x }^{ 2 } } =\frac { 1+\frac { { y }^{ 2 } }{ { x }^{ 2 } }  }{ \frac { y }{ x } -1 } \\ \frac { y }{ x } =t\Rightarrow y=xt\Rightarrow dy=t+xdt\\ t+xdt=\frac { 1+{ t }^{ 2 } }{ t-1 } \\ xdt=\frac { 1+{ t }^{ 2 } }{ t-1 } -t=\frac { 1+{ t }^{ 2 }-{ t }^{ 2 }+t }{ t-1 } =\frac { t+1 }{ t-1 } \\ \int { \frac { t-1 }{ t+1 } dt } =\int { \frac { dx }{ x }  } \\ \int { \left( 1-\frac { 2 }{ t+1 }  \right)  } dt=\int { \frac { dx }{ x }  } \\ t-2\ln { \left| t+1 \right| =\ln { \left| x \right| +C }  } \\ \frac { y }{ x } -2\ln { \left| \frac { y }{ x } +1 \right| =\ln { \left| x \right| +C }  } \\ y\left( 1 \right) =1\\ 1-\ln { 4=C } \\ \frac { y }{ x } -2\ln { \left| \frac { y }{ x } +1 \right| =\ln { \left| x \right| +1- } \ln { 4 }  } \\ \frac { y }{ x } =\ln { \left( { \frac { e }{ 4 } x\left( \frac { x+y }{ x }  \right)  }^{ 2 } \right)  } \\ \\ { \frac { e }{ 4 } x\left( \frac { x+y }{ x }  \right)  }^{ 2 }={ e }^{ \frac { y }{ x }  }\\ { \left( x+y \right)  }^{ 2 }=4{ e }^{ \frac { y }{ x } -1 }\\ \\ $$
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