can someone please check my solution?
Dear noteiler,
You can't obtain, $\displaystyle C_{1}-\frac{1}{p}=\frac{y}{p^{2}}$ by integrating both sides of $\displaystyle \frac{dp}{p^2}=\frac{dy}{p^2}-\frac{2y}{p^3}$. This is because,
$\displaystyle \frac{dp}{p^2}=\frac{dy}{p^2}-\frac{2y}{p^3}dp$
$\displaystyle \int{\frac{dp}{p^2}}=\color{red}{\int{\frac{dy}{p^ 2}}}-\color{red}{\int{\frac{2y}{p^3}dp}}$.
The highlighted integrations could not be evaluated beacause they contain two variables whose relation we don't know.
But instead try,
$\displaystyle (p+2y)\frac{dp}{dy}=p$
Dividing by y, $\displaystyle \left(\frac{p}{y}+2\right)\frac{dp}{dy}=\frac{p}{y }$
Now use the substitution, $\displaystyle u=\frac{p}{y}$
Hope you could countinue from here.
it doesn't work that way, i don't integrate two seperate functions, but i integrate the whole 2-argument function. i mean,
$\displaystyle d(\frac{y}{p^2})=\frac{dy}{p^2}-\frac{2ydp}{p^3}$
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.
$\displaystyle x=2c_{1}p-lnc_{2}{\color{red}\mid{p}\mid}$
$\displaystyle y=c_{1}p^{2}-p$
Hope this will help you.