Question

The complete solution of \( (mz-ny)\frac{\partial z}{\partial x} + (nx-lz)\frac{\partial z}{\partial y} = ly-mx \) is

• A. \( xyz = f(lx+my+nz) \)
• B. \( lx+my+nz = f(x+yz) \)
• C. \( x^2+y^2+z^2 = f(lx+my+nz) \)
• D. \( (lx)^2+(my)^2+(nz)^2 = f(lmnxyz) \)

Hint:

For \(Pp+Qq=R\), use Lagrange's auxiliary equations \(\frac{dx}{P} = \frac{dy}{Q} = \frac{dz}{R}\).
Find two independent solutions \(u(x,y,z)=c_1\) and \(v(x,y,z)=c_2\) using method of multipliers or grouping.
The general solution is then \(F(u,v)=0\) or \(u=f(v)\).

Answer 1

The Correct Option is C

This is Lagrange's linear partial differential equation \(Pp + Qq = R\), where \(p = \partial z / \partial x\) and \(q = \partial z / \partial y\). Here, \(P = mz-ny\), \(Q = nx-lz\), \(R = ly-mx\). The auxiliary equations are \(\frac{dx}{P} = \frac{dy}{Q} = \frac{dz}{R}\): \[ \frac{dx}{mz-ny} = \frac{dy}{nx-lz} = \frac{dz}{ly-mx} \] Choose multipliers \(x, y, z\): Each fraction is equal to \(\frac{x dx + y dy + z dz}{x(mz-ny) + y(nx-lz) + z(ly-mx)}\). Denominator = \(mzx - nxy + nxy - lyz + lyz - mzx = 0\). Thus, \(x dx + y dy + z dz = 0\). Integrating gives \(\frac{x^2}{2} + \frac{y^2}{2} + \frac{z^2}{2} = C_1'\), or \(x^2+y^2+z^2 = c_1\). Let \(u = x^2+y^2+z^2\). Choose multipliers \(l, m, n\): Each fraction is equal to \(\frac{l dx + m dy + n dz}{l(mz-ny) + m(nx-lz) + n(ly-mx)}\). Denominator = \(lmz - lny + mnx - mlz + nly - nmx = 0\). Thus, \(l dx + m dy + n dz = 0\). Integrating gives \(lx + my + nz = c_2\). Let \(v = lx+my+nz\). The complete solution is \(f(u,v)=0\), or \(u = \phi(v)\). So, \(x^2+y^2+z^2 = f(lx+my+nz)\). \[ \boxed{x^2+y^2+z^2 = f(lx+my+nz)} \]