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 equation is a Lagrange linear PDE of the form \(Pp+Qq=R\), where \(z=z(x,y)\), \(p=\partial z/\partial x\), \(q=\partial z/\partial y\), and here \(P=x\), \(Q=y\), \(R=z\).

Step 2: Lagrange's method uses the auxiliary (subsidiary) equations \[\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R} \quad\Rightarrow\quad \frac{dx}{x}=\frac{dy}{y}=\frac{dz}{z}\]

Step 3: Taking the first pair, \(\dfrac{dx}{x}=\dfrac{dy}{y}\). Integrating: \(\ln x=\ln y+c_1\), so \[\frac{x}{y}=c_1\]

Step 4: Taking the second pair, \(\dfrac{dy}{y}=\dfrac{dz}{z}\). Integrating: \(\ln y=\ln z+c_2\), so \[\frac{y}{z}=c_2\]

Step 5: The two independent first integrals are \(x/y=c_1\) and \(y/z=c_2\). The general solution is any arbitrary relation between them: \[\phi\left(\frac{x}{y},\frac{y}{z}\right)=0\] which is option (A). \[\boxed{\phi(x/y,\,y/z)=0}\]