Question

xdy-ydx=square root x2+y2 dx

Answer 1

Please check the attachment. Thank you.

Answer 2

Answer:

ln [sec x + tan x ] = ln x + C

To prove:

$x d y - y d x = \sqrt { x ^ { 2 } + y ^ { 2 } } d x$ is a constant

Solution:

By using the homogenous differential methods,  

$\begin{array} { l } { \frac { d y } { d x } - \frac { y } { x } = \frac { 1 } { x } \times \sqrt { x ^ { 2 } + y ^ { 2 } } } \\ { \frac { d y } { d x } = \frac { y } { x } + \frac { 1 } { x } \times \sqrt { x ^ { 2 } + y ^ { 2 } } } \end{array}$

Let

$\frac {y}{x} = v$

Then

$\frac {dy}{dx}=v+x \frac {dv}{dx}$

$v + x \frac{dv}{dx} = v+\sqrt {\frac{x^2}{x^2} + \frac{y^2}{x^2}}$

$v + x \frac{dv}{dx} = v+\sqrt {1+v^2}$

$\frac{dv}{\sqrt{1+v^2}} =\frac{dx}{x}$

Taking integrals on both sides,

$\int \frac { d v } { \sqrt { 1 + v ^ { 2 } } } = \int \frac { d x } { x }$

Let

$v = \tan x$

$dv = \sec^{2} x dx$

Then

$\begin{aligned} \int \frac { \sec ^ { 2 } x } { \sqrt { 1 + \tan ^ { 2 } x } } d x & = \int \frac { d x } { x } \\ \int \frac { \sec ^ { 2 } x } { \sec x } d x & = \int \frac { d x } { x } \\ \int \sec x d x & = \int \frac { d x } { x } \end{aligned}$

ln [sec x + tan x ] = ln x + C