Turning 2nd order into 1st order

[WelshDave]

Good day all

I am struggling with this one.

m (d^2 h)/(dt^2 )+k dh/dt=-mg

use v=dh/dt in place to turn 2nd order into 1st order

so from this i have got

(d^2 h)/(dt^2 )= -g-(v/m)

Is this a first order equation? If so can somebody point me to solve the equation when m=80, k=8.7, v(0)=0ms-1.

Any pointers/ help would be most welcome.

thanks Dave

 

[MarkFL]

If you are going to use:

$$\frac{dh}{dt}=v$$

then:

$$\frac{d^2h}{dt^2}=\frac{dv}{dt}$$

And your original ODE becomes (after dividing through by $m$):

$$\frac{dv}{dt}+\frac{1}{m}v=-g$$

This is now a first order ODE and you can use an integrating factor to solve.

 

[WelshDave]

I've used the formula

N(y)dy=M(X)dx first order separable ODE

dvdt+(1/m)v=−g

I'm guessing that should be

dv/dt+(k/m)v= -g

k=8.7
m= 80
v(0)=0

worked out a long formula and got,

v(t)= -1/87e^-(87t/800)(7848e^(87t/800)-7848)

Can you advise it that is correct. I looks right to me.

 

[MarkFL]

Okay, we have (I left off $k$ before):

$$\frac{dv}{dt}+\frac{k}{m}v=-g$$

If we multiply through by an integrating factor of:

$$\mu(t)=e^{\frac{kt}{m}}$$

We then obtain:

$$e^{\frac{kt}{m}}\frac{dv}{dt}+\frac{k}{m}e^{\frac{kt}{m}}v=-ge^{\frac{kt}{m}}$$

The LHS may now be rewritten as the derivative of a product:

$$\frac{d}{dt}\left(e^{\frac{kt}{m}}v\right)=-ge^{\frac{kt}{m}}$$

Integrating through w.r.t $t$, there results:

$$e^{\frac{kt}{m}}v=C-\frac{mg}{k}e^{\frac{kt}{m}}$$

Hence:

$$v(t)=Ce^{-\frac{kt}{m}}-\frac{mg}{k}$$

Now, if we take $v(0)=v_0$, we find:

$$C-\frac{mg}{k}=v_0\implies C=v_0+\frac{mg}{k}$$

And so we now have:

$$v(t)=\left(v_0+\frac{mg}{k}\right)e^{-\frac{kt}{m}}-\frac{mg}{k}$$

Now, plugging in for the given constants we find:

$$v(t)=\frac{7840}{87}\left(e^{-\frac{87t}{800}}-1\right)$$

This is close to what you have...you have 7848 where I have 7840. :)