# Thread: Find general solution for given points

1. ## Find general solution for given points

Thanks to all for your help! Please show me where i went wrong. I'm getting zeros as the answer for part a) and b)

Given 2xy(dy/dx) + y^2 - 1 = 0 find all solutions A) (1,0) B) (0,1) in the x-y plane.

Here's my workout:
2xy (dy/dx) +y^2 -1 = 0
(2xy/1-y^2) (dy/dx) =1
(2y/1-y^2) (dy/dx) =1/x

integrate both sides:
S (2y/(1-y^2) (dy/dx) = S 1/x dx
-lnly^2 - 1ldy = ln lxl + C

solve for y get rid of ln take e to both sides
e^(-lnly^2 - 1) = e^(ln lxl + C)

General solution:
y= square root x(e^(C/2))

2. ## Re: Find general solution for given points

Originally Posted by nivek0078
Thanks to all for your help! Please show me where i went wrong. I'm getting zeros as the answer for part a) and b)

Given 2xy(dy/dx) + y^2 - 1 = 0 find all solutions A) (1,0) B) (0,1) in the x-t plane.

Here's my workout:
2xy (dy/dx) +y^2 -1 = 0
(2xy/1-y^2) (dy/dx) =1
(2y/1-y^2) (dy/dx) =1/x

integrate both sides:
S (2y/(1-y^2) (dy/dx) = S 1/x dx
-lnly^2 - 1ldy = ln lxl + C

solve for y get rid of ln take e to both sides
e^(-lnly^2 - 1) = e^(ln lxl + C)

General solution:
y= square root x(e^(C/2))
The x-t plane? Why would you be in the x-t plane when your DE is in terms of x and y?

3. ## Re: Find general solution for given points

I'm sorry it is the x-y plane!

4. ## Re: Find general solution for given points

First of all, when you integrate with the substitution on the LHS, you should end up with \displaystyle \displaystyle \begin{align*} -\ln{\left| 1 - y^2 \right|} \end{align*}

5. ## Re: Find general solution for given points

Yes I have that. When i plug in the points it doesn't generate the right soltuion. can't seem to find where i went wrong in the process.

integrate both sides:
S (2y/(1-y^2) (dy/dx) = S 1/x dx

-ln ly^2 - 1 ldy = ln lxl + C

Then take e to both sides
e^(-lnly^2 - 1) = e^(ln lxl + C)

General solution
y= square root x(e^(C/2))

6. ## Re: Find general solution for given points

Originally Posted by nivek0078
Yes I have that. When i plug in the points it doesn't generate the right soltuion. can't seem to find where i went wrong in the process.

integrate both sides:
S (2y/(1-y^2) (dy/dx) = S 1/x dx

-ln ly^2 - 1 ldy = ln lxl + C

Then take e to both sides
e^(-lnly^2 - 1) = e^(ln lxl + C)

General solution
y= square root x(e^(C/2))
I am quite capable of reading what you wrote. You, however, need to make note of the mistake you made which I pointed out. You have just rewritten exactly the same thing!

7. ## Re: Find general solution for given points

Sorry didn't see it right off. Thank you for the help.