# Thread: separable DEQ

1. ## separable DEQ

I'm getting a incorrect answer for Given 2xy(dy/dx) + y^2 - 1 = 0 find solution A) (1,0) B) (0,1) in x-y plane. Please tell me where i'm screw up and how to fix it.

This is my workout:
2xy(dy/dx) + y^2 - 1

-seperate:
y^2 +2y dy=-2x + 1 dx

-intergrate:
(y^3/3)+(y^2) = x-(x^2) +C
This is where my problem is, I'm not sure from here how to do the following.

-divide through by y
y=square root (-x+C/y+3) I don't think that is right and if it is what is my next step in getting rid of y on the right side?

2. ## Re: separable DEQ

Originally Posted by nivek0078
I'm getting a incorrect answer for Given 2xy(dy/dx) + y^2 - 1 = 0 find solution A) (1,0) B) (0,1) in x-y plane. Please tell me where i'm screw up and how to fix it.

This is my workout:
2xy(dy/dx) + y^2 - 1

-seperate:
y^2 +2y dy=-2x + 1 dx

-intergrate:
(y^3/3)+(y^2) = x-(x^2) +C
This is where my problem is, I'm not sure from here how to do the following.

-divide through by y
y=square root (-x+C/y+3) I don't think that is right and if it is what is my next step in getting rid of y on the right side?
A DE is only separable if you can write it as \displaystyle \displaystyle \begin{align*} f(y)\,\frac{dy}{dx} = g(x) \end{align*}. To properly separate your variables in this case, you would need to do this...

\displaystyle \displaystyle \begin{align*} 2x\,y\,\frac{dy}{dx} + y^2 - 1 &= 0 \\ 2x\,y\,\frac{dy}{dx} &= 1 - y^2 \\ \frac{2x\,y}{1 - y^2}\,\frac{dy}{dx} &= 1 \\ \frac{2y}{1 - y^2} \,\frac{dy}{dx} &= \frac{1}{x} \end{align*}

And now you can integrate both sides...

3. ## Re: separable DEQ

Thank you for your help. I've been doing these problems way to long!

4. ## Re: separable DEQ

Still not getting the correct answer. Please tell me where i went wrong and how to fix it. So i took the intergal as given by Prove It:

-ln(y^2-1) = ln(x) + C

-solve for y have to remove the ln by using e
e^(-ln(y^2-1)) = e^(ln(x) + C) => (1/(y^2)-1) = xe^C

-general solution
y=squart root x(e^C/2)

5. ## Re: separable DEQ

Originally Posted by nivek0078
Still not getting the correct answer. Please tell me where i went wrong and how to fix it. So i took the intergal as given by Prove It:

-ln(y^2-1) = ln(x) + C

-solve for y have to remove the ln by using e
e^(-ln(y^2-1)) = e^(ln(x) + C) => (1/(y^2)-1) = xe^C

-general solution
y=squart root x(e^C/2)
Here is a useful trick with arbitary constants

rewrite $\displaystyle C=\ln(C_2)$

This gives

$\displaystyle -\ln(y^2-1)=\ln(x)+\ln(C_2) \iff \ln(y^2-1)=\ln(C_2x)^{-1} \iff y^2-1=\frac{1}{C_2x} \iff y^2-1=C_3\frac{1}{x}$