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**Ackbeet** I'm studying DE's on my own to fill in gaps in my knowledge. I've come across a DE which, when I follow the procedure, results in a solution that doesn't satisfy the original DE. Here's the DE:

$\displaystyle \dfrac{dy}{dx}=\dfrac{x+3y}{3x+y}.$

The DE is homogeneous, and either substitution ($\displaystyle x=vy$ or $\displaystyle y=ux$) is, by symmetry, the same process. I choose $\displaystyle x=vy$ with $\displaystyle dx=v\,dy+y\,dv.$ Then the DE becomes

$\displaystyle (3x+y)\,dy=(x+3y)\,dx\quad\to\quad(vy+3y)\,(v\,dy+ y\,dv)=(3vy+y)\,dy\quad\to$

$\displaystyle v^{2}y\,dy+3yv\,dy+vy^{2}\,dv+3y^{2}\,dv=3vy\,dy+y \,dy\quad\to\quad(vy^{2}+3y^{2})\,dv=(y-v^{2}y)\,dy$

$\displaystyle \displaystyle\to\quad y^{2}\,(v+3)\,dv=y(1-v^{2})\,dy\quad\to\quad \frac{v+3}{1-v^{2}}\:dv=\frac{dy}{y}\quad\to\quad \ln|y|=\int\frac{v\,dv}{1-v^{2}}+3\int\frac{dv}{1-v^{2}}.$

For the first integral, let $\displaystyle u=1-v^{2}$, and then $\displaystyle du=-2v\,dv,$ or

$\displaystyle -\dfrac{du}{2}=v\,dv,$ and hence we get

$\displaystyle \displaystyle\ln|y|=-\frac{1}{2}\int\frac{du}{u}+\frac{3}{2}\,\ln\left| \frac{v+1}{v-1}\right|+C=

-\frac{1}{2}\ln|1-v^{2}|+\frac{3}{2}\,\ln\left|\frac{v+1}{v-1}\right|+C$

$\displaystyle \displaystyle\to\quad \ln|y^{2}|=\ln\left|\frac{(v+1)^{3}}{(v-1)^{3}}\cdot\frac{1}{1-v^{2}}\right|+C=\ln\left|\frac{(v+1)^{3}}{(v-1)^{3}}\cdot\frac{1}{(1-v)(1+v)}\right|+C=\ln\left|\frac{(v+1)^{2}}{(v-1)^{4}}\right|+C\quad\to$

$\displaystyle \displaystyle y^{2}=C\,\frac{(v+1)^{2}}{(v-1)^{4}}\quad\to\quad y=C\,\frac{v+1}{(v-1)^{2}}\quad\to\quad y=C\,\frac{x/y+1}{(x/y-1)^{2}}=C\,\frac{(x+y)/y}{((x-y)/y)^{2}}=\frac{x+y}{y}\,\frac{y^{2}}{(x-y)^{2}}$

$\displaystyle \displaystyle=\frac{y(x+y)}{(x-y)^{2}}\quad\to\quad y(x-y)^{2}=xy+y^{2}\quad\to\quad (x-y)^{2}=x+y.$

This is our implicit solution. Differentiating with respect to $\displaystyle x$ gives

$\displaystyle \displaystyle 2(x-y)(1-y')=1+y'\quad\to\quad 2(x-y)-1=y'+2y'(x-y)\quad\to\quad 2(x-y)-1=y'(x-y+1)\quad\to$

$\displaystyle \displaystyle y'=\frac{2x-2y-1}{x-y+1}.$

This is not the original DE, nor do I see a method for getting the original DE. I conclude that either my solution is incorrect, or my differentiation is incorrect. Where's my mistake?

Thanks in advance!