Discussion
If the characteristic curves of the partial differential equation \(xu_{xx} + 2x^{2}u_{xy} = u_{x} - 1\) are \(\mu(x, y) = c_{1}\) and \(\nu(x, y) = c_{2}\), where \(c_{1}\) and \(c_{2}\) are constants, then
Answer: A
The characteristics are \(\mu = x^{2} - y\) and \(\nu = y\).
For \(Au_{xx} + 2Bu_{xy} + Cu_{yy} + \cdots = 0\) the characteristic directions satisfy \(A\,dy^{2} - 2B\,dx\,dy + C\,dx^{2} = 0\). Here \(A = x\), \(2B = 2x^{2}\) and \(C = 0\):
\[ \begin{aligned} x\,dy^{2} - 2x^{2}\,dx\,dy &= 0 \\ x\,dy\,(dy - 2x\,dx) &= 0 \end{aligned} \]
So \(dy = 0\), giving \(y = c\), or \(dy = 2x\,dx\), giving \(y - x^{2} = c\), equivalently \(x^{2} - y = c\). The equation is hyperbolic (\(B^{2} - AC = x^{4} \gt 0\) for \(x \neq 0\)), so there are two real families.
\(x^{2} + y\) comes from a sign error in the second family. \(\nu = x^{2}\) would mean \(x\) is constant, \(dx = 0\); putting \(dx = 0\) into the characteristic equation leaves \(x\,dy^{2} = 0\), which forces \(dy = 0\) too, so the lines \(x = \) constant are not characteristics.
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