Question

p sinx + q cosy = tanz through lagranges method​

Answer 1

Lagrange's Solution of a Linear Partial Differential Equation

Solution.

Given,

$\quad \mathrm{p\:sinx+q\:cosy=tanz}$

Lagrange's Auxiliary Equations are

$\quad \mathrm{\frac{dx}{sinx}=\frac{dy}{cosy}=\frac{dz}{tanz}}$

$\Rightarrow \mathrm{cosecx\:dx=secy\:dy=cotz\:dz}$

Taking the first two ratios, we get

$\quad \mathrm{cosecx\:dx=secy\:dy}$

$\Rightarrow \mathrm{\int cosecx\:dx=\int secy\:dy}$

$\Rightarrow \mathrm{ln\left(tan\frac{x}{2}\right)=ln\left\{tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\right\}+ln(C_{1})}$

$\Rightarrow \mathrm{tan\frac{x}{2}=tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\:C_{1}}$

$\Rightarrow \color{blue}\mathrm{C_{1}=\frac{tan\frac{x}{2}}{tan\left(\frac{\pi}{4}+\frac{x}{2}\right)}}$

Taking the last two ratios, we get

$\quad \mathrm{secy\:dy=cotz\:dz}$

$\Rightarrow \mathrm{ln\left\{tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\right\}=ln(sinx)+ln(C_{2})}$

$\Rightarrow \mathrm{tan\left(\frac{\pi}{4}+\frac{x}{2}\right)=sinx\:C_{2}}$

$\Rightarrow \color{blue}\mathrm{C_{2}=\frac{tan\left(\frac{\pi}{4}+\frac{x}{2}\right)}{sinx}}$

$\therefore$ the required general solution is

$\quad\boxed{\color{blue}\mathrm{\phi\left(\frac{tan\frac{x}{2}}{tan\left(\frac{\pi}{4}+\frac{x}{2}\right)},\:\frac{tan\left(\frac{\pi}{4}+\frac{x}{2}\right)}{sinx}\right)=0}}$

where $\mathrm{\phi}$ is any arbitrary function of its arguments.