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Question
(1 + y + x2 y) dx + (x + x3) dy = 0
Solution
\[\left( 1 + y + x^2 y \right)dx + \left( x + x^3 \right)dy = 0\]
\[ \Rightarrow dx + y\left( 1 + x^2 \right)dx + x\left( 1 + x^2 \right)dy = 0\]
\[ \Rightarrow dx + \left( 1 + x^2 \right) \left[ ydx + xdy \right] = 0\]
\[ \Rightarrow \left( 1 + x^2 \right) \left[ ydx + xdy \right] = - dx\]
\[ \Rightarrow \left[ ydx + xdy \right] = - \frac{1}{\left( 1 + x^2 \right)}dx\]
\[ \Rightarrow \left[ ydx + xdy \right] = - \frac{dx}{\left( 1 + x^2 \right)}\]
On integrating both side we get,
\[\left( xy \right) = - \int\frac{1}{1 + x^2}dx\]
\[ \Rightarrow xy = - \tan^{- 1} x + c\]
\[ \Rightarrow xy + \tan^{- 1} x = c\]
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