Discussion
The solution of the differential equation
\[ y^{1} - 2xy = xy^{2} \]
is :
Answer: B
The solution is \(y = \left[-\tfrac{1}{2} + ce^{-x^{2}}\right]^{-1}\).
The \(y^{1}\) printed in the equation is \(y'\), so \(y' - 2xy = xy^{2}\) is a Bernoulli equation. Put \(v = 1/y\), so \(v' = -y'/y^{2}\). Dividing by \(y^{2}\):
\[ \begin{aligned} -v' - 2xv &= x \\ v' + 2xv &= -x \end{aligned} \]
The integrating factor is \(e^{x^{2}}\):
\[ \begin{aligned} (ve^{x^{2}})' &= -xe^{x^{2}} \\ ve^{x^{2}} &= -\tfrac{1}{2}e^{x^{2}} + c \\ v &= -\tfrac{1}{2} + ce^{-x^{2}} \end{aligned} \]
So \(y = 1/v\). The squared forms do not invert \(v\), and \(e^{+x^{2}}\) comes from the wrong sign of the integrating factor.
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