# Thread: Is my solution correct?

1. ## Is my solution correct?

can someone please check my solution?

2. Originally Posted by noteiler
can someone please check my solution?
Dear noteiler,

Can you please tell me how you obtained the equation shown below,

$C_{1}-\frac{1}{p}=\frac{y}{p^{2}}$

3. Originally Posted by Sudharaka
Dear noteiler,

Can you please tell me how you obtained the equation shown below,

$C_{1}-\frac{1}{p}=\frac{y}{p^{2}}$
by integrating both sides of
$\frac{1}{p^2}=\frac{dy}{p^2} - \frac{2dpy}{p^3}$

4. Originally Posted by noteiler
by integrating both sides of
$\frac{1}{p^2}=\frac{dy}{p^2} - \frac{2dpy}{p^3}$
Dear noteiler,

You can't obtain, $C_{1}-\frac{1}{p}=\frac{y}{p^{2}}$ by integrating both sides of $\frac{dp}{p^2}=\frac{dy}{p^2}-\frac{2y}{p^3}$. This is because,

$\frac{dp}{p^2}=\frac{dy}{p^2}-\frac{2y}{p^3}dp$

$\int{\frac{dp}{p^2}}=\color{red}{\int{\frac{dy}{p^ 2}}}-\color{red}{\int{\frac{2y}{p^3}dp}}$.

The highlighted integrations could not be evaluated beacause they contain two variables whose relation we don't know.

$(p+2y)\frac{dp}{dy}=p$

Dividing by y, $\left(\frac{p}{y}+2\right)\frac{dp}{dy}=\frac{p}{y }$

Now use the substitution, $u=\frac{p}{y}$

Hope you could countinue from here.

5. Originally Posted by Sudharaka
Dear noteiler,

You can't obtain, $C_{1}-\frac{1}{p}=\frac{y}{p^{2}}$ by integrating both sides of $\frac{dp}{p^2}=\frac{dy}{p^2}-\frac{2y}{p^3}$. This is because,

$\frac{dp}{p^2}=\frac{dy}{p^2}-\frac{2y}{p^3}dp$

$\int{\frac{dp}{p^2}}=\color{red}{\int{\frac{dy}{p^ 2}}}-\color{red}{\int{\frac{2y}{p^3}dp}}$.

The highlighted integrations could not be evaluated beacause they contain two variables whose relation we don't know.
it doesn't work that way, i don't integrate two seperate functions, but i integrate the whole 2-argument function. i mean,
$d(\frac{y}{p^2})=\frac{dy}{p^2}-\frac{2ydp}{p^3}$
left side is the complete differential or whatever it is called in English, so we can integrate it. of course, there is a propability that i am wrong or making silly mistake, but i use this very often and so far i has been correct

6. Originally Posted by noteiler
it doesn't work that way, i don't integrate two seperate functions, but i integrate the whole 2-argument function. i mean,
$d(\frac{y}{p^2})=\frac{dy}{p^2}-\frac{2ydp}{p^3}$
left side is the complete differential or whatever it is called in English, so we can integrate it. of course, there is a propability that i am wrong or making silly mistake, but i use this very often and so far i has been correct
You're OK with what you've done except at the very end. THe solution I got was

$x = 2 c_1 p - \ln p + c_2$
$y = c_1p^2 - p$.

7. Originally Posted by Danny
You're OK with what you've done except at the very end. THe solution I got was

$x = 2 c_1 p - \ln p + c_2$
$y = c_1p^2 - p$.
thanks for confirming, i've already started to worry
but isn't that wright
$\ln p + c_2 = \ln p + \ln c_2 = \ln c_2p$

8. Originally Posted by noteiler
thanks for confirming, i've already started to worry
but isn't that wright
$\ln p + c_2 = \ln p + \ln c_2 = \ln c_2p$
You're right. Point well taken.

9. Originally Posted by noteiler
it doesn't work that way, i don't integrate two seperate functions, but i integrate the whole 2-argument function. i mean,
$d(\frac{y}{p^2})=\frac{dy}{p^2}-\frac{2ydp}{p^3}$
left side is the complete differential or whatever it is called in English, so we can integrate it. of course, there is a propability that i am wrong or making silly mistake, but i use this very often and so far i has been correct
Dear noteiler,

Thank you very much for this post. I learnt something new because of your thread. Although your answer is almost correct you are missing out a important modulus sign.

$x=2c_{1}p-lnc_{2}{\color{red}\mid{p}\mid}$

$y=c_{1}p^{2}-p$