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My answer is: $[(1+x^2)^3]y = \dfrac{(1+x^2)^3}3+C$
But this option is not given, so is it correct?


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I got the same answer as you. I don't understand why they haven't got our one in the options. I solved it by multiplying through by an integrating factor. Namely, $$ \exp \left( \displaystyle \int \dfrac{6x}{1+x^2} \text{ d}x \right)$$ – Kari Sep 1 '14 at 1:16
The answer should be $a$. – Mhenni Benghorbal Sep 1 '14 at 2:16
I think that your solution is equivalent to $(a)$. Try expanding out the $(1+x^2)^3$ part. – Akiva Weinberger Sep 1 '14 at 4:13
I see thanks for the help – Lim Zhi Jian Sep 1 '14 at 4:47

$$(1+x^2) \frac{dy}{dx}=2x-6xy \Rightarrow (1+x^2) \frac{dy}{dx}=2x(1-3y) \Rightarrow \frac{dy}{1-3y}=\frac{2x}{1+x^2} dx \\ \Rightarrow -\frac{1}{3} \ln |1-3y|=\ln |1+x^2|+c$$

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Thanks for the help. – Lim Zhi Jian Sep 1 '14 at 4:47

$$(1+x^2)\frac{dy}{dx}+6xy=2x \Rightarrow (1+x^2)\frac{dy}{dx}=2x-6xy \Rightarrow (1+x^2)\frac{dy}{dx}=2x(1-3y) \Rightarrow \frac{1}{1-3y}dy=\frac{2x}{1+x^2}dx \Rightarrow \int \frac{1}{1-3y}dy=\int \frac{2x}{1+x^2}dx \\ \Rightarrow \frac{-1}{3}\ln{|1-3y|}=\ln{|1+x^2|}+c\Rightarrow \ln{|1-3y|}^{\frac{-1}{3}}=\ln{|1+x^2|}+c \\ \Rightarrow e^{\ln{|1-3y|}^{\frac{-1}{3}}}=e^{\ln{|1+x^2|}+c } \\ \Rightarrow |1-3y|^{\frac{-1}{3}}=C|1+x^2|, C=e^c$$

Solve now for $y$.

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After the third implication it should be $1+x^2$ I guess – Alexander Frei Sep 1 '14 at 1:22
You're right! I edited my answer.. – Mary Star Sep 1 '14 at 1:51
Thanks for the help. – Lim Zhi Jian Sep 1 '14 at 4:53

$$\begin{align} (1+x^2)\frac{dy}{dx}+6xy&=2x\\ (1+x^2)\frac{dy}{dx}&=2x(1-3y)\\ \int \frac{dy}{1-3y}&=\int\frac {2x}{1+x^2}dx\\ -\frac13\ln(1-3y)&=\ln(1+x^2)\\ (1-3y)(1+x^2)^3&=0\\ (1+x^2)^3&=3y(1+x^2)^3\\ 1+3x^2+3x^4+x^6&=3y(1+x^2)^3\\ \frac 13+x^2+x^4+\frac{x^6}3&=y(1+x^2)^3\\ y(1+x^2)^3&=x^2+x^4+\frac{x^6}3+C \end{align}$$

i.e. option (A) $\blacksquare$.

Check by differentiating:

$$\begin{align} (1+x^2)^3\frac{dy}{dx}+6xy(1+2x^2)^2&=2x+4x^3+2x^5\\ &=2x(1+2x^2+x^4)\\ &=2x(1+x^2)^2\\ (1+x^2)\frac{dy}{dx}+6xy&=2x \end{align}$$

which is the original equation, hence solution is correct.

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Thanks for the help – Lim Zhi Jian Sep 1 '14 at 4:46
you're most welcome :) – hypergeometric Sep 1 '14 at 4:48

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