Rewrite the equation to $\frac{dx}{dy}+\frac{x}{y+1}=\frac{-y}{y+1}$

I can use used the integrating factor $μ(y)=e^{-\int\frac{1}{y+1}dy}$ to solve it.

The answer is $x=\frac{-y^2}{2(y+1)}+\frac{C}{y+1}$

I'm curious that are there any other methods to solve this ODE. I am asking for a different method because sometimes I am not able(or haven't enough time, this one costs me about half an hour to solve it ) to rewrite the equation into a proper form. So, if I could obtain other methods to solve this kind of ODEs, I may have more chances to solve it in a exam.

  • $\begingroup$ it should be $y(x)=x\pm \sqrt{x2+2x+C}$$ $\endgroup$ – Dr. Sonnhard Graubner Nov 27 '15 at 15:22

See that we can write as $$(x+y)dy+(y+1)dx=0$$ $$xdy+ydx+ydy+dx=0$$ $$d(xy)+ydy+dx=0$$ now just integrate.

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    $\begingroup$ Nice catch:) +1 $\endgroup$ – Rowan Nov 27 '15 at 15:21
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    $\begingroup$ Plain and straightforward as long as one can see it. +1 $\endgroup$ – Shailesh Nov 27 '15 at 16:42

Notice, the given D.E. can be easily solved by exact differential form as follows $$(y+1)dx+(x+y)dy=0$$ Now, comparing the above equation with $Mdx+Ndy=0$, we get $$M=y+1\implies \frac{\partial M}{\partial y}=1$$ $$N=x+y\implies \frac{\partial N}{\partial x}=1$$ since, $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$, hence the given equation is in the exact differential form, hence the solution is given as $$\int_{\text{keeping y constant}} (y+1)dx+\int_{\text{terms free of}\ x}(x+y)dy=C$$ $$ (y+1)\int dx+\int y\ dy=C$$ $$\color{blue}{(y+1)x+\frac{y^2}{2}=C}$$

  • $\begingroup$ What if the case $\frac{\partial M}{\partial y}\ne \frac{\partial N}{\partial x}$, then we can't use this method anymore? For example,$2ydy-2ydx+xdy=0$ $\endgroup$ – Rowan Nov 27 '15 at 15:45
  • $\begingroup$ Use Integration factor as follows $$\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}=f(x)$$ or $$\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}=f(y)$$ $\endgroup$ – Harish Chandra Rajpoot Nov 27 '15 at 15:49
  • $\begingroup$ I use the first factor but it gives me $(-2-\frac{x}{y})dx+2dy=0$. Am I did it wrong? And does $$\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}=f(x)$$ and $$\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}=f(y)$$ a general factor that I can always use when$\frac{\partial M}{\partial y}\ne \frac{\partial N}{\partial x}$ $\endgroup$ – Rowan Nov 27 '15 at 16:11
  • $\begingroup$ Well explained. +1. $\endgroup$ – Shailesh Nov 27 '15 at 16:41

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