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I have a problem and I got most of the solution, but don't understand how to proceed. The problem is to solve:

$$(x+2y)y'=1, \qquad y(0)=-1.$$

Here is my reasoning:

Substitute $z = x+2y$. Then $z'=1+2y'$. So $y'=\frac{z'-1}{2}$. So $\frac{z'-1}{2} = \frac{1}{z}$. So $z' = \frac{2+z}{z}$. So $\frac{dz}{dx}$ ($x$ taken without loss of generality) = $\frac{2+z}{z}$. So $\frac{zdz}{2+z} = dx$. So $z - 2\ln|z+2| = x+C$. Therefore, $$(x+2y) - 2\ln|x+2y+2| = x+C.$$

Am I correct that there is no member of this one-parameter family of solutions that satisfies the initial condition $y(0)=-1$, because plugging in does not work?

Also, is there a singular solution, and how to find it??

Thanks.

nmasanta
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  • Formally, when one tries to satisfy the initial conditions, one finds $C=\infty$. This hints what is the form of the singular solution - the only way to have this is to set $y=-1-\frac{x}{2}$. – Start wearing purple Aug 25 '13 at 14:00
  • How did you get y = -1 - x/2? What about y = -x/2, because we divide by (x+2y) on both sides –  Aug 25 '13 at 14:08
  • This corresponds to zero of the logarithm. Actually, it is better to write your general solution in the form (after exponentiating) $$|x+2y+2|=\tilde{C}e^{y},$$ where I introduce instead of $C$ another constant of integration $\tilde{C}=e^{-C/2}$. The solution you are interested in corresponds to $\tilde{C}=0$. – Start wearing purple Aug 25 '13 at 14:12
  • @O.L. Sorry, could you clarify your comment a little? What corresponds to zero of the logarithm? And also, do we just consider the term that we divide by (for the singular solution) when separating variables? –  Aug 25 '13 at 14:26
  • The solution $y=-1-x/2$ appears when $\ln |x+2y+2|$ diverges. The division by $x+2y$ does not lead to any problem, since $x+2y=0$ does not verify the initial conditions and therefore can not solve your problem. But I feel all these words are a bit misleading. You should just understand why the formula in my previous comment is equivalent to your general solution, and then to determine the constant $\tilde{C}$ from the initial conditions. – Start wearing purple Aug 25 '13 at 14:32

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