1. ## separable equation

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.

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

2. Originally Posted by yaykittyeee
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.

$\displaystyle \frac{dy}{dx}=\sqrt{y}cos^2\sqrt{y}$
If you post all your work then what you're doing wrong can be easily pointed out to you.

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

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

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

4. Originally Posted by yaykittyeee
$\displaystyle \frac{1}{\sqrt{y}cos^2\sqrt{y}}dy=dx$
$\displaystyle 2ln(cos^2\sqrt{y})=x+C$
Your integration is wrong (differentiate it .... you don't get back the integrand). The correct approach would begin by making the substitution $\displaystyle u = \sqrt{y}$.