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I tried to solve: $$\displaystyle xyy'-y^2= (x+y)^2 \cdot e^{\frac{-y}{x}} \ \ \ \ \ \ \ \ \ \ \ (1)$$ I took: $y(x)= x\cdot z(x) \Longrightarrow y'(x)=z'(x)+z(x)$, and I substituted in $(1)$: $$z(x)\cdot z'(x) \cdot x = e^{-z}\cdot (1+2z(x)+[z(x)]^2 $$ Using separation of variables: $$\frac{e^{z(x)} z(x)z'(x)}{(1+z(x))^2}=\frac{1}{x} \ \ \ \ \ \ \ \ \ \ \ (2) $$ Let be $1+z(x)= w(x)$, (2) became $$e^{w(x)-1} [w(x)]^{-2}w'(x)=\frac{1}{x}$$ Integrating this: $$\int e^{w(x)-1} [w(x)]^{-2}w'(x) \ dx=\int \frac{1}{x} \ dx= \log |x|+C$$ But I can't integrate

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if $y(x)=xz(x)$, then $y'(x)=xz'(x)+z(x)$, not what you wrote. –  Mariano Suárez-Alvarez Oct 7 '11 at 4:34
    
ok, i can see it... –  Mario De León Urbina Oct 7 '11 at 6:33
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2 Answers 2

up vote 2 down vote accepted

$y=zx \Rightarrow y'=z'x+z$

$xzx(z'x+z)-z^2x^2=(x+zx)^2e^\frac{-zx}{x}$

$zx^2(z'x+z)-z^2x^2=(x+zx)^2e^{-z}$

$z'zx^3=(x+zx)^2e^{-z}$

$z'zx=(1+z)^2e^{-z}$

$$\frac{ze^z}{(1+z)^2} dz=\frac{1}{x}dx \Rightarrow \int\frac{ze^z}{(1+z)^2}dz=\int\frac{1}{x}dx $$ ,The first integral we can solve using integration by parts where $u=e^z$ ,and $dv=\frac{z}{(1+z)^2}dz$. Solution for second integral is obvious.

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Before $(2)$ I can see that all it's right, except for the derivation over $y(x)$, but the substitution it's not wrong, because I forgot the factor $x$ that Mariano and Anon pointed. I was realized the resolution in paper. Anon, certainly was a bad idea the introduction of $w(x)$, and @pedja rightly jumped from $(2)$. –  Mario De León Urbina Oct 7 '11 at 6:26
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Your first mistake was in applying the product rule. Observe: $$y'=(xz)'=x'z+xz'=1\cdot z(x)+x\cdot z(x)\ne z(x)+z'(x).$$ Your second mistake was when you substituted $z+z'$ (even though that's wrong) for $y$: $$x(xz)(z+z')-(xz)^2=(x^2+2x(xz)+(xz)^2)e^{-z}\quad(\text{cancel, divide by }x^2)$$ $$\implies zz'=(1+2z+z^2)e^{-z}.$$ (Which has one less $x$ in it.) Your third mistake was in substituting $z=w-1$; you should've got $$\frac{e^{w-1}(w-1)w'}{w^2}=\frac{1}{x}.$$

Finally - if your derivation thus far would have been correct, though it isn't - you should have used the fact that $dw=w'(x)dx$ to change the left-hand integral thusly: $$\int e^{w-1}w^{-2}dw=\log|x|+C. $$


Also, hint: notice that dividing both sides by $xy$ gives $$y'-\frac{y}{x}=\left(1+\frac{y}{x}\right)\left(\frac{x}{y}+1\right)\exp\left(-\frac{y}{x}\right).$$ Do you know how to solve differential equations of the form $y'=F(y/x)$?

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Yeah, @anon: take $z= \frac{y}{x}$... –  Mario De León Urbina Oct 7 '11 at 6:32
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