D.E. (2x+1)dy + (y^2)dx = 0 through (4,1)

[xoninhas]

Find the solution to the diff. eq.:

(2x+1)dy + (y^2)dx = 0

that goes through point (4,1)...

I just know that I should separate both y and x so I got:

dy/y^2= - dx/2x+1

what now?... :S I don't know how the integration in this case goes...

 

[royhaas]

Integrate both sides and add a constant. Then use the initial conditions to evaluate the constant.

 

[xoninhas]

to which side do I add a constant?? then I solve towards the constant?
Thanks for the time.

 

[xoninhas]

Ok, so integrating I got this: (i finally managed to integrate it right I think...)

- 1/y = - ln(2x+1)/2

where does the constant go?

then I get e^(2/y) = (2x +1) but I should get e^(2/y) = C(2x +1) where does C come from.

 

[galactus]

You are OK separating variables.

\(\displaystyle \int\frac{1}{y^{2}}dy+\int\frac{1}{2x+1}dx=0\)

\(\displaystyle \frac{-1}{y}=\frac{-ln(2x+1)}{2}+C\)

Now, we can use the IC to find C, x=4 and y=1:

\(\displaystyle \frac{-1}{1}=\frac{-ln(2(4)+1)}{2}+C\)

\(\displaystyle C=ln(3)-1\)

Using C and solving for y we get:

\(\displaystyle \boxed{y=\frac{2}{ln(2x+1)-2(ln(3)-1)}}\)