# Integrating Factor

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

$y -xy' = x + yy'$

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

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

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 $tan^{-1} (\frac{x}{y})$ and $-ln\sqrt{x^2 + y^2}$ come ?
• Mar 17th 2008, 06:58 AM
Peritus
first divide the ODE by x

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

now let us use the following substitution:

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

after applying this substitution to our ODE we get:

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

this is a separable equation:

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

$

\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...