1. ## Separation of variables..

I get stuck at the integration step

use the method of separation of variables to find the solution of the differential equation:

dy/dx=sqrt(2y-y^2)/x+1 , y(0)=1

2. Originally Posted by Raidan
I get stuck at the integration step

use the method of separation of variables to find the solution of the differential equation:

dy/dx=sqrt(2y-y^2)/x+1 , y(0)=1
So $\displaystyle \int \frac{dy}{\sqrt{2y-y^2}} = \int \frac{dx}{x+1}$
then complete the square in the square root term

$\displaystyle \sqrt{2y-y^2} = \sqrt{1-(y-1)^2}$ Let $\displaystyle u = y-1$ and you should be able to integrate.

3. could someone help me to solve this question?

4. Originally Posted by Raidan
could someone help me to solve this question?
Danny pointed it out:

u = y-1 so du = dy

$\displaystyle \int{\frac{du}{\sqrt{1-u^2}}} = \int{\frac{dx}{x+1}}$
The left side is a standard integral and is equal to $\displaystyle arcsin(u) +A$
For the right side recall that $\displaystyle \int{\frac{f'(x)}{f(x)}} = ln|f(x)| + B$