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General solution of differential equation: $r \sin \theta d\theta +(r^3 -2 r^2 \cos \theta +\cos \theta)dr=0$

$(a)\;r^2(1+ce^{-r})=2 \sin \theta \\ (b)\;r^2(1+ce^{-r})=2 \cos \theta \\ (c)\;r(1+ce^{-r^2})=2 \cos \theta \\ (d)\;r(1+ce^{-r})=2 \sin \theta $
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$r \sin \theta \large\frac{d\theta}{dr}$$ +\cos \theta (1-2r^2)=-r^3$
$\cos \theta=t; -\sin \theta \large\frac{d \theta}{dr}=\large\frac{dt}{dr}$
$- r \large\frac{dt}{dr}$$ +t(1-2r^2)=-r^3$
$\large\frac{dt}{dr}$$+t (2r -\large\frac{1}{r})=r^2$
$I.F= e^{\int (r- \large\frac{1}{r} )dr}$
$\quad= e^{r^2- \ln r}=\large\frac{e^{r^2}}{r}$
$\large\frac{e^{r^2}}{r}t =\int \large\frac{e^{r^2}}{r}$$ (-r^2)dr$
$\large\frac{-e^{r^2}}{r} =\int r e^{r^2} $$dr$
$r^2 =t$
$2rdr=dt$
$-\large\frac{e^r}{r}.$$ \cos \theta =\large\frac{1}{2} \int e^t.dt$
$-\large\frac{e^r}{r}.$$ \cos \theta =\large\frac{1}{2} e^t.+c$
$-\large\frac{e^r}{r}.$$ \cos \theta =\large\frac{1}{2}$$ e^{r^2}+c$
$-\large\frac{e^r}{r}.$$ \cos \theta +\large\frac{e^{r^2}}{2}$$=c$
$ r(1+ce^{-r^2})=2 \cos \theta$
Hence c is the correct answer.
answered Feb 4, 2014 by meena.p
 

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