Separation of variables..

**Raidan** · May 4th 2009, 04:24 AM

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

**Danny** · May 4th 2009, 04:44 AM

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 $\int \frac{dy}{\sqrt{2y-y^2}} = \int \frac{dx}{x+1}$
then complete the square in the square root term

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

**Raidan** · May 4th 2009, 06:30 AM

could someone help me to solve this question?

**e^(i*pi)** · May 4th 2009, 06:36 AM

Originally Posted by Raidan

could someone help me to solve this question?

Danny pointed it out:

u = y-1 so du = dy

Therefore your equation becomes

$\int{\frac{du}{\sqrt{1-u^2}}} = \int{\frac{dx}{x+1}}$

The left side is a standard integral and is equal to $arcsin(u) +A$
wikipedia link (and don't forget to put y-1 wherever u appears)

For the right side recall that $\int{\frac{f'(x)}{f(x)}} = ln|f(x)| + B$

Where A and B are constants and so C = B-A which is also a constant (your constant of integration)

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