2
$\begingroup$

$2r(1+s^2)dr + (1+r^4)ds = 0$

$\Rightarrow \frac {2r}{1+r^4}dr + \frac {ds}{1+s^2} = 0$ $(1)$

Integrate both sides:

For integration of $\frac {2r}{1+r^4}dr$ $(2)$

I put $1+r^4=t\Rightarrow 4r^3dr=dt\Rightarrow 2rdr = \frac {dt}{2r^2} = \frac {dt}{2\sqrt{t-1}}$

Hence $(2)$ becomes:

  • Integration of $\frac {dt}{2\sqrt{t-1}}$ = $\sqrt{t-1} = \sqrt{r^4} = r^2$
  • Integration of $\frac {ds}{1+s^2} = arctan(s)$

Hence (1) becomes

$$r^2 + arctan(s) = c$$

But at the back, the answer written is $r^2 + s = c(1-r^2s)$

Can someone please tell me what did I do wrong.

  • 0
    is here $$r=r(s)$$ or $$s=s(r)$$? – 2017-02-08
  • 1
    The $dr$ term is $$\frac{2rdr}{1+r^4}=\frac{dt}{2t\sqrt{t-1}}$$ instead of $$\frac{dt}{2\sqrt{t-1}}$$ hence one should solve $$\int\frac{dt}{2t\sqrt{t-1}}+\int\frac{ds}{1+s^2}=C$$ An alternative is to use the change of variable $$u=r^2$$ which yields $$\frac{2rdr}{1+r^4}=\frac{du}{1+u^2}$$ hence $$\arctan u+\arctan s=C$$ and maybe you can finish from this? – 2017-02-08
  • 1
    $$\int\frac {2r}{1+r^4}dr=\int\frac {2r}{1+(r^2)^2}dr=\arctan r^2$$ so **(1)** --> $$\arctan r^2+\arctan s=C$$ – 2017-02-08
  • 0
    @Dr.SonnhardGraubner Another garbage comment? – 2017-02-08
  • 0
    @MyGlasses thanks – 2017-02-08
  • 0
    To continue the preceding comments. one can have $$\arctan r^2+\arctan s=\arctan \frac{r^2+s}{1-r^2s}$$ and so the solution is $$\frac{r^2+s}{1-r^2s}=c,$$ where $c=\tan C$ – 2017-02-08
  • 1
    You know the derivative of $r^2$ is $2r$. That integration result should have raised a giant red flag. – 2017-02-08
  • 1
    Also, if you don't know @ΘΣΦGenSan's formula, you can always just use the formula for $\tan(a+b)$. – 2017-02-08
  • 0
    @Mike Are you saying that my comment is wrong? – 2017-02-08
  • 1
    @ΘΣΦGenSan I'm saying that the formula for $\tan(a+b)$ is more commonly taught than the formula for $\tan^{-1}a+\tan^{-1}b$. – 2017-02-08
  • 0
    @Mike Ah I see. Nevermind – 2017-02-08

1 Answers 1

1

In addition to @Did's hint, there is a relatively easy way to handle that integral.

We have $$ \frac {2r}{1+ r^4} \mathrm {d}r = - \frac {1}{1+ s^2 } \mathrm {d}s $$ $$\Rightarrow \int \frac {\mathrm {d}(r^2)}{1+(r^2)^2} = \int - \frac {1}{1+ s^2 } \mathrm {d}s$$ $$\Rightarrow \arctan r^2 = - \arctan s +c $$

Hope you can take it from here.

  • 0
    "In addition to" = "Merely repeating"? – 2017-02-12