Question

The general solution of the linear partial differential equation (PDE): \(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=z\) is ____ (\(\phi\) is an arbitrary function).

• A. \(\phi(x/y,y/z)=0\)
• B. \(\phi(x^2-z^2,x^3-y^3)=0\)
• C. \(\phi(y/z,x^2+y^2+z^2)=0\)
• D. \(\phi(x+y+z,xyz)=0\)

Hint:

Use Lagrange's auxiliary equations dx/x = dy/y = dz/z and integrate in pairs.

Answer 1

The Correct Option is A

Step 1: The subsidiary equations $\dfrac{dx}{x}=\dfrac{dy}{y}=\dfrac{dz}{z}$ describe characteristic curves. Introduce a parameter $s$ along each curve so that \[\frac{dx}{ds}=x,\qquad \frac{dy}{ds}=y,\qquad \frac{dz}{ds}=z\]

Step 2: Each is a simple linear ODE with solution \[x=x_0e^{s},\qquad y=y_0e^{s},\qquad z=z_0e^{s}\] where $(x_0,y_0,z_0)$ is the starting point of the characteristic at $s=0$.

Step 3: Since all three coordinates carry the same exponential factor $e^{s}$, their ratios stay fixed along every characteristic: \[\frac{x}{y}=\frac{x_0}{y_0}=\text{constant}, \qquad \frac{y}{z}=\frac{y_0}{z_0}=\text{constant}\]

Step 4: So $x/y$ and $y/z$ are two functionally independent invariants of the characteristic system (the characteristics are straight lines through the origin in $(x,y,z)$-space).

Step 5: The general solution of the PDE is therefore an arbitrary relation between these invariants: \[\phi\left(\frac{x}{y},\frac{y}{z}\right)=0\] confirming option (A). \[\boxed{\phi(x/y,\,y/z)=0}\]