# Thread: General solution of differential equation - question 2

1. ## General solution of differential equation - question 2

I have the question

2) Find the general solution of the following differential equation:

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

i have fount the general solution to be:

re arrange>> 1/y^2 -1 dy = -1/x dx
integrate>> 1/2 ln(y-1) - 1/2 ln(y+1) = lnx + C

is that correct?? Or can i simplify further??

2. ## Re: General solution of differential equation - question 2

Originally Posted by NFS1
I have the question

2) Find the general solution of the following differential equation:

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

i have fount the general solution to be:

re arrange>> 1/y^2 -1 dy = -1/x dx
integrate>> 1/2 ln(y-1) - 1/2 ln(y+1) = lnx + C

is that correct?? Or can i simplify further??
Dear NFS1,

Your differential equation is not clear. Do you mean,

$\frac{dy}{dx}=\frac{y^2-1}{x}\mbox{ or }\frac{dy}{dx}=y^2-\frac{1}{x}$ ?

3. ## Re: General solution of differential equation - question 2

Originally Posted by Sudharaka
Dear NFS1,

Your differential equation is not clear. Do you mean,

$\frac{dy}{dx}=\frac{y^2-1}{x}\mbox{ or }\frac{dy}{dx}=y^2-\frac{1}{x}$ ?
For the OP, if it's the first, then it's separable. If it's the second, it's a Bernoulli equation.

4. ## Re: General solution of differential equation - question 2

Originally Posted by Sudharaka
Dear NFS1,

Your differential equation is not clear. Do you mean,

$\frac{dy}{dx}=\frac{y^2-1}{x}\mbox{ or }\frac{dy}{dx}=y^2-\frac{1}{x}$ ?
Its the first one mate...i have seperated and got my final answer as...1/2 ln(y-1) - 1/2 ln(y+1) = lnx + C

5. ## Re: General solution of differential equation - question 2

What you wrote looks fine, but don't forget the modulus signs where necessary.
Now you have to determine the general solution ...

Can you do that? ...

6. ## Re: General solution of differential equation - question 2

what i wrote does it simplify to >> ln (y-1/y+1) = ln x + C ??
i dont quite understand what to do from here??

7. ## Re: General solution of differential equation - question 2

Yes, that's correct and write $C=\ln(e^C)$ (you don't need modulus signs here because $e^C>0$, so you get:
$\ln\left|\frac{y-1}{y+1}\right|=2\ln|x|+2\ln(e^C)$
$\Leftrightarrow \ln\left|\frac{y-1}{y+1}\right|= \ln\left(x^2\cdot e^{2C}\right)$
$\Leftrightarrow \left|\frac{y-1}{y+1}\right|=x^2\cdot e^{2C}$
$\Leftrightarrow \left|1-\frac{2}{y+1}\right|=x^2\cdot e^{2C}$
$\Leftrightarrow y=...$

Try to finish it ...

8. ## Re: General solution of differential equation - question 2

y = (-1/x^2.e^2C) - 1 ???

9. ## Re: General solution of differential equation - question 2

I would say:
$1-\frac{2}{y+1}=\pm x^2 \cdot e^{2C}$, let $\pm e^{2C}=K$
$\Leftrightarrow \frac{2}{y+1}=1-(x^2\cdot K)$
$\Leftrightarrow y+1=\frac{2}{1-(x^2\cdot K)}$
$\Leftrightarrow y=\frac{2}{1-(x^2\cdot K)}-1$
$\Leftrigharrow y= \frac{2-(1-x^2\cdot K)}{1-x^2\cdot K}$
$\Leftrightarrow y=\frac{1+x^2\cdot K}{1-x^2\cdot K}$

10. ## Re: General solution of differential equation - question 2

Originally Posted by Siron
I would say:
$1-\frac{2}{y+1}=\pm x^2 \cdot e^{2C}$, let $\pm e^{2C}=K$
$\Leftrightarrow \frac{2}{y+1}=1-(x^2\cdot K)$
$\Leftrightarrow y+1=\frac{2}{1-(x^2\cdot K)}$
$\Leftrightarrow y=\frac{2}{1-(x^2\cdot K)}-1$
$\Leftrigharrow y= \frac{2-(1-x^2\cdot K)}{1-x^2\cdot K}$
$\Leftrightarrow y=\frac{1+x^2\cdot K}{1-x^2\cdot K}$
Just a quick question why does ... y-1/y+1 become 1-2/y+1 ??

11. ## Re: General solution of differential equation - question 2

Because:
$1-\frac{2}{y+1}=\frac{y+1}{y+1}-\frac{2}{y+1}=\frac{y+1-2}{y+1}=\frac{y-1}{y+1}$