Help with the Separation of Variables and Integration

[erobz]

looks like we moved that drag force over to the right hand side

I meant examine similarities between post 16 and post 32.

 

[silento]

. however -2B is in place of the mass

 

[erobz]

View attachment 340974 . however -2B is in place of the mass

luckily, that's just a constant... Quite literally solve for ##v \frac{dv}{dx}## in terms of ##\frac{du}{dx}## in post 32, and plug it into post 16 equation. Don't forget about the ##u## substitution made in 16 either. If you do it correctly, watch the equation transform into a butterfly.

I 've got to try to get to bed though. Good luck.

 

[silento]

 

[erobz]

View attachment 340975

Try again. everything should be in terms of ##u,x##..no more ##v## in the equation.

 

[silento]

From post 32

then I plugged that into the answer from post 16 in place of vdv/dx. I don't see where I'm going wrong

 

[erobz]

From post 32 View attachment 340976 then I plugged that into the answer from post 16 in place of vdv/dx. I don't see where I'm going wrong

in post 16 eqn, substitute on the LHS to get rid of ##v## and teh constant... remember I defined ##u = \varphi - \beta v^2##.

So you end with a simple ODE in terms of variable ##u## and ##x##.

 

[silento]

Oh right!

 

[silento]

wait a minute. I think I see where you are going with this

 

[silento]

correct me in i'm wrong, but we are going to move the du to the LHS. Then we can sub v back in for u. Then velocity would be the only variable on the LHS along with the constants. On the right hand side it would just be 1/-2Bdx which is just constants???

 

[erobz]

correct me in i'm wrong, but we are going to move the du to the LHS. Then we can sub v back in for u. Then velocity would be the only variable on the LHS along with the constants. On the right hand side it would just be 1/-2Bdx which is just constants???

Solve for ##u(x)## from the resulting ODE. After you’ve done that you can change back to ##v## and it’s values etc…

 

[silento]

Im sorry but what's ODE?

 

[erobz]

Im sorry but what's ODE?

Ordinary differential equation. The equation we just developed. Linear. separable, ODE.

 

[silento]

solve for u(x)? I guess I was wrong. I was thinking I could just do

. Then I can plug u back in and change u to v. So, all the v would be on the left while the x is on the right. I can then take the integral of both sides and solve for xf-xi

 

[silento]

(1/2) is to the power. After I get u(x) then I change it back to v then do I integrate? 120 to 0 is bounds for velocity and I need to find ∆x which is just xf-xi which are the bounds for dx(position)

 

[silento]

but there is two answers. a + and a - since we were taking the square root

 

[silento]

?

 

[erobz]

The equation you need to solve is:

$$\frac{- 2 \beta}{m} u = \frac{du}{dx} $$

Separate variable's and integrate.

And if you don't reply in Latex math formatting I'm calling it quits with a clear conscience...
See: LaTeX Guide

 

[silento]

\begin{equation}
\ln|u| = -\frac{2B}{m}x
\end{equation}

 

[erobz]

\begin{equation}
\ln|u| = -\frac{2B}{m}x
\end{equation}

Integrate ##u## with limits ##u_f, u_o##. You can leave the ##x## the way it is assuming ##x_o=0##. Otherwise good.

 

[silento]

so let me think about this. u is velocity from 54 m/s to 0 m/s. x_f is what I'm solving for. Would I only integrate the LHS?

 

[erobz]

so let me think about this. u is velocity from 54 m/s to 0 m/s. x_f is what I'm solving for. Would I only integrate the LHS?

##u## is not ##v##. Go back and look at what ##u## is…

 

[erobz]

The basic equation is fine. But it’s missing a constant of integration. That is why I suggest you “re do it” integrating it as a definite integral between limits ##u_f## and ##u_o##.

 

[silento]

\begin{equation}
\ln|u_f| - \ln|u_o| = -\frac{2B}{m}x
\end{equation}

 

[erobz]

\begin{equation}
\ln|u_f| - \ln|u_o| = -\frac{2B}{m}x
\end{equation}

Now combine the logs

 

[silento]

\begin{equation}
\ln\left|\frac{u_f}{u_o}\right| = -\frac{2B}{m}x
\end{equation}

 

[erobz]

\begin{equation}
\ln\left|\frac{u_f}{u_o}\right| = -\frac{2B}{m}x
\end{equation}

exponentiate both sides, and move ##u_o## over to the right hand side.

 

[silento]

\left|\frac{u_f}{u_o}\right| = e^{-\frac{2B}{m}x}

 

[silento]

am I typing something wrong?

 

[erobz]

am I typing something wrong?

Delimiters? And you can lose the absolute value bars. The RHS is always positive.

 

[silento]

\begin{equation}\left| \frac{u_f}{u_o} \right| = e^{-\frac{2B}{m}x}\end{equation}

 

[erobz]

\begin{equation}\left| \frac{u_f}{u_o} \right| = e^{-\frac{2B}{m}x}\end{equation}

The absolute value bars are no longer necessary. The RHS is always positive. Move ##u_o## over.

 

[silento]

\begin{equation}u_f = u_o \cdot e^{-\frac{2B}{m}x}\end{equation}

 

[erobz]

\begin{equation}u_f = u_o \cdot e^{-\frac{2B}{m}x}\end{equation}

Now evaluate ##u_f## and ##u_o## using the corresponding ##v_f## and ##v_o##. Make sure you are evaluating ##u##. I repeat ##u## is not ##v##.

 

[erobz]

Now evaluate ##u_f## and ##u_o## using the corresponding ##v_f## and ##v_o##. Make sure you are evaluating ##u##. I repeat ##u## is not ##v##.

Oh I just realized you are trying to solve for ##x##, not ##v##. Anyhow just finish it out, the solve for ##x##. I realized this is undoing a few steps. My bad.