Help with the Separation of Variables and Integration

[silento]

hold on I think I got something

 

[erobz]

du= -2Bvdv

I was looking for this:

$$ \frac{du}{dx} = - 2 \beta v \frac{dv}{dx} $$

Which is what basically what you wrote (Don't drop the independent variable from the notation).

 

[silento]

Alright...that makes sense. Interested to see where we go from here

 

[erobz]

Alright...that makes sense. Interested to see where we go from here

look at the RHS of that result, and compare it to the RHS of your original equation. Notice anything?

 

[silento]

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

 

[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 the 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