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?
Thanks
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Akiva Weinberger
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Lim Zhi Jian
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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, $$ \exp \left( \displaystyle \int \dfrac{6x}{1+x^2} \text{ d}x \right)$$ – Khallil Sep 01 '14 at 01:16
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The answer should be $a$. – Mhenni Benghorbal Sep 01 '14 at 02:16
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1I think that your solution is equivalent to $(a)$. Try expanding out the $(1+x^2)^3$ part. – Akiva Weinberger Sep 01 '14 at 04:13
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I see thanks for the help – Lim Zhi Jian Sep 01 '14 at 04:47
3 Answers
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$$(1+x^2) \frac{dy}{dx}=2x-6xy \Rightarrow (1+x^2) \frac{dy}{dx}=2x(1-3y) \Rightarrow \frac{dy}{1-3y}=\frac{2x}{1+x^2} dx \\ \Rightarrow -\frac{1}{3} \ln |1-3y|=\ln |1+x^2|+c$$
evinda
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$$(1+x^2)\frac{dy}{dx}+6xy=2x \Rightarrow (1+x^2)\frac{dy}{dx}=2x-6xy \Rightarrow (1+x^2)\frac{dy}{dx}=2x(1-3y) \Rightarrow \frac{1}{1-3y}dy=\frac{2x}{1+x^2}dx \Rightarrow \int \frac{1}{1-3y}dy=\int \frac{2x}{1+x^2}dx \\ \Rightarrow \frac{-1}{3}\ln{|1-3y|}=\ln{|1+x^2|}+c\Rightarrow \ln{|1-3y|}^{\frac{-1}{3}}=\ln{|1+x^2|}+c \\ \Rightarrow e^{\ln{|1-3y|}^{\frac{-1}{3}}}=e^{\ln{|1+x^2|}+c } \\ \Rightarrow |1-3y|^{\frac{-1}{3}}=C|1+x^2|, C=e^c$$
Solve now for $y$.
Mary Star
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$$\begin{align} (1+x^2)\frac{dy}{dx}+6xy&=2x\\ (1+x^2)\frac{dy}{dx}&=2x(1-3y)\\ \int \frac{dy}{1-3y}&=\int\frac {2x}{1+x^2}dx\\ -\frac13\ln(1-3y)&=\ln(1+x^2)\\ (1-3y)(1+x^2)^3&=0\\ (1+x^2)^3&=3y(1+x^2)^3\\ 1+3x^2+3x^4+x^6&=3y(1+x^2)^3\\ \frac 13+x^2+x^4+\frac{x^6}3&=y(1+x^2)^3\\ y(1+x^2)^3&=x^2+x^4+\frac{x^6}3+C \end{align}$$
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.
Hypergeometricx
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