Solving $$xy' = y + e^{\frac{y}{x}}$$ after I sub $u = y/x$ into the equation, I don't know how to solve integral of $u$, is it just $0.5u^2$ or I should consider $x$ and $y$?
How to solve $xy' = y + e^{\frac{y}{x}}$ by first making change of $u = y/x$
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calculus
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0How do you mean just consider $y,x$ do you mean you want to solve and transform back into $x$ and $y$? If I was a teacher I would prefer to transform back as there are a bunch of transforms that one could do for some equations (not in this case) so I would rather mark one correct answer than multiple transformed ones. – 2017-02-10
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0I don't know when to solve directly and when to sub into a third variable. – 2017-02-10
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$$x{ y }^{ \prime }=y+{ e }^{ \frac { y }{ x } }\\ \\ y=xu\Rightarrow \quad { y }^{ \prime }=u+x{ u }^{ \prime }\\ xu+{ x }^{ 2 }{ u }^{ \prime }=xu+{ e }^{ u }\\ { x }^{ 2 }{ u }^{ \prime }={ e }^{ u }\\ \int { { e }^{ -u }du } =\int { \frac { dx }{ { x }^{ 2 } } } \\ -{ e }^{ -u }=-\frac { 1 }{ x } +C\\ { e }^{ -\frac { y }{ x } }=\frac { 1 }{ x } +C\\ y=-x\ln { \left| \frac { 1 }{ x } +C \right| } \\ \\ y=x\ln { \left| \frac { x }{ 1+xC } \right| } $$
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0today I made so idiotic mistakes, thank you @Chinny84 – 2017-02-10
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0haha - you are talking to the master of...idiotic mistakes! – 2017-02-10
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0thank you again,for showing me my mistakes) – 2017-02-10
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0I can say reason of your false. Your answer needs more sign ⇔ or some explanation. These are not "good proof without words". Although you can only solve some species of homogeneous differential equation, I notice that your answer without words cause your false. – 2017-02-11