Question

solve the p.d.e (mz-ny)p+(nx-lz)q=ly-mx​

Answer 1

Partial Differential Equation

Solution. The given partial differential equation is

$\quad (mz-ny)p+(nx-lz)q=ly-mx$

Then, Lagrange's subsidiary equations are

$\quad\frac{dx}{mz-ny}=\frac{dy}{nx-lz}=\frac{dz}{ly-mx}\quad...(i)$

Step 1.

We take multipliers x, y, z to the ratios of (i) respectively. Then

$\quad\frac{x\:dx+y\:dy+z\:dz}{(mz-ny)x+(nx-lz)y+(ly-mx)z}$

$\quad=\frac{x\:dx+y\:dy+z\:dz}{0}$

This gives: $x\:dx+y\:dy+z\:dz=0$

On integration, we get

$\quad x^{2}+y^{2}+z^{2}=c_{1}$ where $c_{1}$ is integral constant.

Step 2.

Now we take multipliers l, m, n to the ratios of (i) respectively. Then

$\quad\frac{l\:dx+m\:dy+n\:dz}{l(mz-ny)+m(nx-lz)+n(nx-lz)}$

$\quad=\frac{l\:dx+m\:dy+n\:dz}{0}$

This gives: $l\:dx+m\:dy+n\:dz=0$

On integration, we get

$\quad lx+my+nz=c_{2}$ where $c_{2}$ is integral constant.

Answer step. Hence the required general solution is $\quad\phi(x^{2}+y^{2}+z^{2},\:lx+my+nz)=0$, where $\phi$ is an arbitrary function.