## Astrophysics one year ago Err...what?

1. Astrophysics

If we have $\frac{ dp }{ dt } = \frac{ p-900 }{ 2 } \implies \frac{ dp/dt}{p-900 }= \frac{ 1 }{ 2 }$ so how does this go to $\frac{ d }{ dt }\ln|p-900|=1/2$

2. Astrophysics

I mean we can do a u - sub, but I find that weird for differentiation

3. Astrophysics

I guess it involves the chain rule, but I'm not seeing it xD

4. Astrophysics

I should specify p=p(t)

5. zzr0ck3r

$$\int\dfrac{1}{p-900}dp=\int\dfrac{1}{2}dt\\\ln(p-900)=\dfrac{1}{2}t+c$$

6. Astrophysics

Yeah that's fine I just mean how to go from $\frac{ dp/dt }{ p-900 }=1/2 \implies \frac{ d }{ dt } \ln|p-900|=1/2$

7. Astrophysics

This is the step before integration

8. Astrophysics

Though I would do it your way, I'm just trying to figure out how to use the chain rule here, if that makes sense

9. zzr0ck3r

I have no idea... it makes no sense to me I did $\dfrac{dp}{dt}\dfrac{1}{p-900}=\dfrac{1}{2}\\\implies \dfrac{1}{p-900}dp=\dfrac{1}{2}dt$ Then I integrate both sides.

10. Astrophysics

Yeah exactly, that's how I'd do it, but in my book it says "By the chain rule the left side" then they show above

11. zzr0ck3r

$$\frac{ d }{ dt } \ln|p-900|=1/2 \implies \dfrac{df(p)}{dt}=\dfrac{1}{2}$$ which is odd to me.

12. zzr0ck3r

the derivative of a function of p with respect to t is a constant 1/2 ?

13. Astrophysics

Ok so this is part of a modelling problem it starts off as $\frac{ dp }{ dt } = \frac{ p-900 }{ 2 } \implies \frac{ dp/dt }{ p-900 } = \frac{ 1 }{ 2 } (eq 6)$ then it says by the chain rule the left side of the equation 6 is the derivative of ln|p-900| with respect to t, so we have $\frac{ d }{ dt }\ln|p-900|=1/2$ then they integrate

14. zzr0ck3r

I don't know man, maybe someone else will help. I suck at calculus...

15. Astrophysics

Haha, no worries man, I was like I should know this...but I had no clue, thanks a lot for the help though!

16. zzr0ck3r

:(

17. Astrophysics

The solution is the same as yours I just don't know why it's written this way

18. zzr0ck3r

Do they end with $$p=Ce^{\frac{1}{2}t}+900$$?

19. Astrophysics

Yup

20. zzr0ck3r

yeah, I dont know what that step is about then.

21. Astrophysics

Yeah neither do I haha, thanks for checking though!

22. freckles

$\frac{d}{dt} \ln|p-900| \text{ does equal } (p-900)' \cdot \frac{1}{p-900} =p' \cdot \frac{1}{p-900}=\frac{dp}{dt} \cdot \frac{1}{p-900}$ but yeah I would have never written it like the way they did

23. Astrophysics

Yeah so they sort of our just going backwards? Just a weird way to introduce dif equations I thought haha, thanks @freckles

24. anonymous

You do stuff like that plenty of time when solving linear ODEs. Without going into too much detail, consider the general case: $\frac{dy}{dx}+f(x)y=g(x)$ You find some function $$\mu(x)$$ which you distribute onto both sides, giving $\mu(x)\frac{dy}{dx}+f(x)\mu(x)y=g(x)\mu(x)$ with the property that $\frac{d}{dx}\left[\mu(x)y\right]=\mu(x)\frac{dy}{dx}+f(x)\mu(x)y$ This means you can solve what remains, i.e. $\frac{d}{dx}\left[\mu(x)y\right]=g(x)\mu(x)$ by simply integrating, i.e. $\mu(x)y=\int g(x)\mu(x)\,dx~~\implies~~y=\frac{1}{\mu(x)}\int g(x)\mu(x)\,dx$

25. anonymous

PS: "Without going into too much detail..." by which I mean I'm assuming you haven't tackled ODEs with methods beyond separating variables.

26. zzr0ck3r

loved it

27. anonymous

28. zzr0ck3r

linear operators at the rescue

29. Astrophysics

Ha, nice! Thanks @SithsAndGiggles And yeah I just started ODE xD

30. Astrophysics

Thanks @zzr0ck3r @freckles @SithsAndGiggles