Answer to Question #202730 in Differential Equations for Kevin

Question

Solve yzdx-xzdy-(x^2+y^2)tan^-1(y/x)dz=0

Answer 1

Question:-

$yzdx-xzdy-(x^2+y^2)tan^{-1}(\frac{y}{x})dz=0..........(1)$

Given

$yzdx-xzdy-(x^2+y^2)tan^{-1}(\frac{y}{x})dz=0..........(1)\\or\\ yzdx-xzdy=(x^2+y^2)tan^{-1}(\frac{y}{x})dz$

Lagrange's auxiliary equation

$\frac{dx}{yz}=\frac{dy}{-xz}=\frac{dz}{(x^2+y^2)tan^{-1}(\frac{y}{x})}$

$from \space I=II$

-xdx=ydy

y²+x²=c₁............(2)

again

$yzdx-xzdy-(x^2+y^2)tan^{-1}(\frac{y}{x})dz=0\\ or \\ ydx-xdy-(x^2+y^2)tan^{-1}(\frac{y}{x})\frac{dz}{z}=0$

let

$\\ \frac{y}{x}=t\\ diffrentiate\\ \frac{xdy-ydx}{x^2}=dt$

put the values and we get

$(-x^2dt)-(x^2+y^2)tan^{-1}(t)\frac{dz}{z}=0\\ -dt -(1+t^2) tan^{-1}(t) \frac{dz}{z}=0\\ or\\ \frac{dt}{(1+t^2)(tan^{-1}(t))}+\frac{dz}{z}=0\\ let \\tan^{-1}(t)=m\\ differentiate\\ \frac{dt}{(1+t^2)}=dm\\ put \space values,\\ \frac{dm}{m}+\frac{dz}{z}=0\\ integrate\\ log(mz)=log(c_2)\\ mz=c_2\\ again \space put \space values(m,t)\\ z .tan^{-1}(\frac{y}{x})=c_2........(3) \\hence\\ answer\\ f(c_1,c_2)=0 \\f(x^2+y^2,z.tan^{-1}(\frac{y}{x})=0$