# Simple differential equation problem

• Apr 6th 2009, 03:59 PM
chengbin
Simple differential equation problem
$\displaystyle (1+x^2)y\frac {dy} {dx}+(1+y^2)x=0$

This is the steps the solution book gave.

Rearranging the original equation

$\displaystyle \frac {y} {1+y^2} dy=-\frac {x} {1+x^2} dx$

Integrating both sides

$\displaystyle \frac {1} {2} \ln(1+y^2)=-\frac {1} {2} \ln(1+x^2)+c$

Usually in this step I solve for y (which is what the book has been doing before too), and I get rid of the ln function. But in this question my solution book gives this.

$\displaystyle \ln[(1+x^2)(1+y^2)]=2c$

$\displaystyle (1+x^2)(1+y^2)=e^{2c}$

Why?

$\displaystyle (1+x^2)(1+y^2)=k$ after replacing k with $\displaystyle e^{2c}$

Is this the same thing as solving for y, but written in a different form?
• Apr 6th 2009, 04:06 PM
mr fantastic
Quote:

Originally Posted by chengbin
$\displaystyle (1+x^2)y\frac {dy} {dx}+(1+y^2)x=0$

This is the steps the solution book gave.

Rearranging the original equation

$\displaystyle \frac {y} {1+y^2} dy=-\frac {x} {1+x^2} dx$

Integrating both sides

$\displaystyle \frac {1} {2} \ln(1+y^2)=-\frac {1} {2} \ln(1+x^2)+c$

Usually in this step I solve for y (which is what the book has been doing before too), and I get rid of the ln function. But in this question my solution book gives this.

$\displaystyle \ln[(1+x^2)(1+y^2)]=2c$

$\displaystyle (1+x^2)(1+y^2)=e^{2c}$

Why?

$\displaystyle (1+x^2)(1+y^2)=k$ after replacing k with $\displaystyle e^{2c}$