# calculus - population model/de

• May 11th 2008, 01:29 AM
mathshelpbook
calculus - population model/de
having so much trouble with de's in general! i think it's a linear de, but really stuck! please help:

dp/dt = -50 + 15p - p^2

• May 11th 2008, 02:03 AM
flyingsquirrel
Hello

There is a $\displaystyle p^2$ in the expression so the equation is not linear. However, the two variables $\displaystyle p$ and $\displaystyle t$ can be separated : $\displaystyle \frac{\mathrm{d}p}{p^2-15p+50}=-\mathrm{d}t$ and we'll get the solution by integrating the both sides.

To achieve this, one needs to make appear a derivative on the LHS. As the denominator contains a $\displaystyle p^2$, one can try to transform $\displaystyle \frac{\mathrm{d}p}{p^2-15p+50}$ into $\displaystyle \mathrm{constant}\cdot\frac{\mathrm{d}u}{1+u^2}$ which anti-derivative is $\displaystyle \mathrm{constant}\cdot \arctan u$. Hence solving this problem boils down to finding $\displaystyle u$. Do you have any idea on how this can be done ?
• May 11th 2008, 02:26 AM
mathshelpbook
thanks flyingsquirrel.
well i have no idea, but i thought maybe i could try 'flipping' it as the rhs is in terms of p

ie. dt/dp = - 1/(p^2 - 15p + 50)
and then t = - integral[1/(p^2 - 15p + 50)] dp

does that work? i'm not sure if that is how to start the question