$\displaystyle

xyy'=\frac{x^2}{cos(\frac{y}{x})}+y^2

$

i know i need to look for a variable which both its derivative and him selft are present in the equation.

i tried t=y/x

but its derivative is not present in the equation

??

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- Aug 17th 2009, 03:11 AMtransgalactichow to pick a variable here??
$\displaystyle

xyy'=\frac{x^2}{cos(\frac{y}{x})}+y^2

$

i know i need to look for a variable which both its derivative and him selft are present in the equation.

i tried t=y/x

but its derivative is not present in the equation

?? - Aug 17th 2009, 01:12 PMMatt Westwood
This appears to be homogeneous if you divide all thru by xy.

Then you'll have a function in y/x. Make the substitution z = y/x, substitute for dy/dx (it's z + x dz/dx) and you will find you have a diffeq with separable variables.