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solve by linear DE: \[(2xy + x^2 + x^4)dx - (1+x^2)dy = 0\]
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i think this is solveable by exact DE easily...but my problem is if it can be solved by linear DE..if so how?
It is exact. However it can also be solved by linear DE if you know how to solve for \(y' + p(x).y = q(x) \) (which indirectly uses the concept of exactness)
how can i put it in that form??
\((2xy+x^{2}+x^{4})dx = (1+x^{2})dy \) \[\Large \Rightarrow \frac{dy}{dx} = \frac{2x}{1+x^{2}}y + x^{2}\]\[\Rightarrow \Large y' - \frac{2x}{1+x^{2}}y = x^{2}\] where \(\Large p(x) = \frac{-2x}{1+x^{2}}, q(x) = x^{2} \)
OHHHH I SEE IT NOW!!!! (although you forgot + x^4 there)
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oh nevermind
i saw what you did
nice ..incredible
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