# Integrating Factor

• Mar 17th 2008, 06:19 AM
Altair
Integrating Factor
The question is :-

$\displaystyle y -xy' = x + yy'$

The question asks to find the integrating factor and to solve it further.

As this is a homogeneous equation I tried the formula of $\displaystyle \frac{1}{Mx + Ny}$ and both partial fraction formulas but the answer which is $\displaystyle \frac{1}{x^2 + y^2}$ wouldn't come. I get $\displaystyle \frac{1}{-x^2-y^2}.$

Secondly even by using the answer of I.F I get stuck at final integration. Please do it for me. I dunno that how does the terms $\displaystyle tan^{-1} (\frac{x}{y})$ and $\displaystyle -ln\sqrt{x^2 + y^2}$ come ?
• Mar 17th 2008, 06:58 AM
Peritus
first divide the ODE by x

$\displaystyle \begin{gathered} \frac{y} {x} - y' = 1 + \frac{y} {x}y' \hfill \\ \hfill \\ \Leftrightarrow y' = \frac{{\frac{y} {x} - 1}} {{\frac{y} {x} + 1}} \hfill \\ \end{gathered}$

now let us use the following substitution:

$\displaystyle v = \frac{y} {x} \Rightarrow y' = v + xv'$

after applying this substitution to our ODE we get:

$\displaystyle xv' = - \frac{{1 + v^2 }} {{1 + v}}$

this is a separable equation:

$\displaystyle \int {\frac{1} {x}dx} = - \int {\frac{{1 + v}} {{1 + v^2 }}} dv$

$\displaystyle \ln \left| x \right| + C = - \arctan (v) - 0.5\ln \left( {1 + v^2 } \right)$

you can now back substitute to obtain the solution in an implicit form.
• Mar 17th 2008, 07:19 AM
Altair
I want to do it by using the I.F.
• Mar 17th 2008, 07:31 AM
Peritus
Quote:

Originally Posted by Altair
I want to do it by using the I.F.

be my guest...