Please explainhow did he simplified this given?

**davewhite123** · January 31st 2010, 03:14 AM

Hi, Can you please show me the step-by-step process of this example. Especially the second and third line?

Here is the given:
$(xy-x)dx=(x^2y^2-x^2)+(y^2-1)dy$
$x(y-1)dx=(x^2(y^2-1)+(y^2-1))dy$
$=(x^2+1)(y^2-1)dy$
$xdx=(x^2+1)(y+1)dy$

Any help will be appreciated, thanks

**Archie Meade** · January 31st 2010, 05:11 AM

Originally Posted by davewhite123

Hi, Can you please show me the step-by-step process of this example. Especially the second and third line?

Here is the given:
$(xy-x)dx=(x^2y^2-x^2)+(y^2-1)dy$
$x(y-1)dx=(x^2(y^2-1)+(y^2-1))dy$
$=(x^2+1)(y^2-1)dy$
$xdx=(x^2+1)(y+1)dy$

Any help will be appreciated, thanks

$(xy-x)dx=(x^2y^2-x^2)+(y^2-1)dy$

x is a common factor of xy and x, so $xy-x=x(y-1)$

$x^2$ is a common factor of $x^2y^2$ and $x^2$ , so $x^2y^2-x^2=x^2(y^2-1)$

Now $y^2-1$ is a common factor, so $x^2(y^2-1)+(y^2-1)=(y^2-1)(x^2+1)$

Finally, $y^2-1=(y+1)(y-1)$

hence y-1 is common to both sides and can be eliminated to obtain the last line.

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