Discussion
The differential equation :
\[ (x^{3} + xy^{4})\, dx + 2y^{3}\, dy = 0 \]
will become exact on multiplication by :
Answer: A
The integrating factor is \(e^{x^{2}}\).
With \(M = x^{3} + xy^{4}\) and \(N = 2y^{3}\), the equation is not exact, since \(M_{y} = 4xy^{3}\) and \(N_{x} = 0\). Test for an integrating factor depending on x alone:
\[ \frac{M_{y} - N_{x}}{N} = \frac{4xy^{3}}{2y^{3}} = 2x \]
This depends only on x, so \(\mu = e^{\int 2x\, dx} = e^{x^{2}}\). Check: after multiplying, \(\partial_{y}(e^{x^{2}}M) = 4xy^{3}e^{x^{2}}\) and \(\partial_{x}(2y^{3}e^{x^{2}}) = 4xy^{3}e^{x^{2}}\), which agree.
\(e^{x}\) and \(e^{-x}\) correspond to the ratio being \(\pm 1\), not \(2x\), and \(e^{x^{2}} + x\) is not of the form \(e^{\int f(x) dx}\) at all.
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