separable equation

yaykittyeee · #1 $Old$ November 16th, 2009, 07:01 PM

could someone please help me with this one? my answer doesn't match the one in the book so i know i'm doing something wrong but i'm not sure where.

$\frac{dy}{dx}=\sqrt{y}cos^2\sqrt{y}$

mr fantastic · #2 $Old$ November 16th, 2009, 08:08 PM

Quote:

Originally Posted by yaykittyeee $View Post$

could someone please help me with this one? my answer doesn't match the one in the book so i know i'm doing something wrong but i'm not sure where.

$\frac{dy}{dx}=\sqrt{y}cos^2\sqrt{y}$

If you post all your work then what you're doing wrong can be easily pointed out to you.

yaykittyeee · #3 $Old$ November 16th, 2009, 10:55 PM

i start with the equation above, then i separated the dy/dx
$\frac{1}{\sqrt{y}cos^2\sqrt{y}}dy=dx$

then i integrate both sides and get
$2ln(cos^2\sqrt{y})=x+C$

that's as far as i get. i'm not sure that i'm integrating the left side right.

mr fantastic · #4 $Old$ November 17th, 2009, 02:49 AM

Quote:

Originally Posted by yaykittyeee $View Post$

i start with the equation above, then i separated the dy/dx
$\frac{1}{\sqrt{y}cos^2\sqrt{y}}dy=dx$

then i integrate both sides and get
$2ln(cos^2\sqrt{y})=x+C$

that's as far as i get. i'm not sure that i'm integrating the left side right.

Your integration is wrong (differentiate it .... you don't get back the integrand). The correct approach would begin by making the substitution $u = \sqrt{y}$ .