I've been looking over some papers on the Method of Characteristics but I've found something I'm not sure about. For quasi-linear equations of the form $$a(x,y,u)u_x + b(x,y,u)u_y = c(x,y,u)$$ I have found two methods.

One involves writing $$\frac{\mathrm dx}{a(x,y,u)} = \frac{\mathrm dy}{b(x,y,u)} = \frac{\mathrm du}{c(x,y,u)}$$ and then combining pairs of the three terms to produce two differentials that are equal to zero and thus represent characteristics.

Thus, for the equation $$x u_x + y u_y = 2xy \quad u = 2 \, \text{on} \, y=x^2$$ we write

$$\frac{\mathrm dx}{x} = \frac{\mathrm dy}{y} = \frac{\mathrm du}{2xy}$$

and then an easy starter for ten is

\begin{align*}

\frac{\mathrm dx}{x} = \frac{\mathrm dy}{y} \implies \frac{\mathrm dx}{x} - \frac{\mathrm dy}{y} &= 0 \\

y\,\mathrm dx - x\,\mathrm dy &= 0 \implies \mathrm d (\tfrac{y}{x}) = 0

\end{align*}

For the second, we must use the $\mathrm du$ term getting, for example

\begin{align*}

\frac{\mathrm dx}{x} = \frac{\mathrm du}{2xy} \implies \frac{\mathrm dx}{x} - \frac{\mathrm du}{2xy} &= 0 \\

2xy\,\mathrm dx - x\,\mathrm du &= 0

\end{align*}

Now, I need to get $\mathrm d(xy-u)=0$ (in order to match the other method which gives a solution that works in the equation). The only way I can see to get there is to write

\begin{align*}

2xy\,\mathrm dx - x\,\mathrm du &= 0 \\

2y\,\mathrm dx - \mathrm du &= 0 \\

2x\tfrac{y}{x}\,\mathrm dx - \mathrm du &= 0 \\

\left(2x\tfrac{y}{x}\,\mathrm dx + x^2\mathrm d (\tfrac{y}{x})\right) - \mathrm du &= 0 \\

\mathrm d(x^2\tfrac{y}{x}) - \mathrm du &= 0 \\

\mathrm d(xy - u) &= 0

\end{align*}

Is that correct? Does that mean that I can never say that I'm holding variables constant in this process? That is, because $y$ is a variable in the equation it must be treated as one even though its differential doesn't appear in the equation?

My solution is $$u(x,y) = xy + 2 - \frac{y^3}{x^3}$$

One involves writing $$\frac{\mathrm dx}{a(x,y,u)} = \frac{\mathrm dy}{b(x,y,u)} = \frac{\mathrm du}{c(x,y,u)}$$ and then combining pairs of the three terms to produce two differentials that are equal to zero and thus represent characteristics.

Thus, for the equation $$x u_x + y u_y = 2xy \quad u = 2 \, \text{on} \, y=x^2$$ we write

$$\frac{\mathrm dx}{x} = \frac{\mathrm dy}{y} = \frac{\mathrm du}{2xy}$$

and then an easy starter for ten is

\begin{align*}

\frac{\mathrm dx}{x} = \frac{\mathrm dy}{y} \implies \frac{\mathrm dx}{x} - \frac{\mathrm dy}{y} &= 0 \\

y\,\mathrm dx - x\,\mathrm dy &= 0 \implies \mathrm d (\tfrac{y}{x}) = 0

\end{align*}

For the second, we must use the $\mathrm du$ term getting, for example

\begin{align*}

\frac{\mathrm dx}{x} = \frac{\mathrm du}{2xy} \implies \frac{\mathrm dx}{x} - \frac{\mathrm du}{2xy} &= 0 \\

2xy\,\mathrm dx - x\,\mathrm du &= 0

\end{align*}

Now, I need to get $\mathrm d(xy-u)=0$ (in order to match the other method which gives a solution that works in the equation). The only way I can see to get there is to write

\begin{align*}

2xy\,\mathrm dx - x\,\mathrm du &= 0 \\

2y\,\mathrm dx - \mathrm du &= 0 \\

2x\tfrac{y}{x}\,\mathrm dx - \mathrm du &= 0 \\

\left(2x\tfrac{y}{x}\,\mathrm dx + x^2\mathrm d (\tfrac{y}{x})\right) - \mathrm du &= 0 \\

\mathrm d(x^2\tfrac{y}{x}) - \mathrm du &= 0 \\

\mathrm d(xy - u) &= 0

\end{align*}

Is that correct? Does that mean that I can never say that I'm holding variables constant in this process? That is, because $y$ is a variable in the equation it must be treated as one even though its differential doesn't appear in the equation?

My solution is $$u(x,y) = xy + 2 - \frac{y^3}{x^3}$$

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