Non linear 1st order differential

H.Bisho18

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Hello
So i have this question. I use F=ma where a is the derivative of v with respect to t, and i get what looks like a first order differential but it's v2 and i'm not sure how to do that. Or if I'm meant to solve the question this way as it seems too high level, any ideas?

thanks
1578779673604.png
 

Jomo

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looks like a first order differential but it's v2 and i'm not sure how to do that.
What do you mean by this?
Can we see your work so we can help you? You would have received help by now if you followed the guidelines and shared your work with us.
 

H.Bisho18

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looks like a first order differential but it's v2 and i'm not sure how to do that.
What do you mean by this?
Can we see your work so we can help you? You would have received help by now if you followed the guidelines and shared your work with us.
sorry heres my working. I get tried making a first order differential which i couldn't solve and then tried a second order which i know is wrong20200111_230754.jpg
 

MarkFL

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I'd begin with the IVP:

\(\displaystyle \d{v}{t}=6000-15v^2=15(20^2-v^2)\) where \(v(0)=5\)

The ODE is separable, and may be written in the following differential form:

\(\displaystyle \frac{1}{20^2-v^2}\,dv=15\,dt\)

What do we get upon applying a partial fraction decomposition on the LHS?
 

Jomo

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sorry heres my working. I get tried making a first order differential which i couldn't solve and then tried a second order which i know is wrongView attachment 16003
When you divide the lhs of the equal sign by 200 you need to do the same to the rhs!
 

H.Bisho18

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I'd begin with the IVP:

\(\displaystyle \d{v}{t}=6000-15v^2=15(20^2-v^2)\) where \(v(0)=5\)

The ODE is separable, and may be written in the following differential form:

\(\displaystyle \frac{1}{20^2-v^2}\,dv=15\,dt\)

What do we get upon applying a partial fraction decomposition on the LHS?
Ah yes i see now thank you!
However I did ignore +c when I intergrated as I couldn't see how you would determine it since the question doesnt say it starts at rest or anything.
Doing this I get the answer in the mark scheme but I was wondering if you know how you'd know c is 0 or was it just an oversight in the question?

20200112_134242.jpg
 

Subhotosh Khan

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Ah yes i see now thank you!
However I did ignore +c when I intergrated as I couldn't see how you would determine it since the question doesnt say it starts at rest or anything.
Doing this I get the answer in the mark scheme but I was wondering if you know how you'd know c is 0 or was it just an oversight in the question?

View attachment 16014
1578844772800.png
I do not see anywhere that "c" (I assume the c = constant of integration) was assumed to be zero - in response #5.
 

MarkFL

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Ah yes i see now thank you!
However I did ignore +c when I intergrated as I couldn't see how you would determine it since the question doesnt say it starts at rest or anything.
Doing this I get the answer in the mark scheme but I was wondering if you know how you'd know c is 0 or was it just an oversight in the question?
Ah, yes, I failed to use Newton's 2nd law...the ODE should in fact be:

\(\displaystyle 3000\d{v}{t}=6000-15v^2\)

\(\displaystyle 200\d{v}{t}=20^2-v^2\)

I would continue as follows:

\(\displaystyle 5\left(\frac{1}{v+20}-\frac{1}{v-20}\right)\,dv=dt\)

Integrate, using the boundaries as limits and switch the dummy variables:

\(\displaystyle 5\int_5^v \frac{1}{y+20}-\frac{1}{y-20}\,dy=\int_0^t\,dz\)

\(\displaystyle 5\ln\left(\frac{3}{5}\left|\frac{v+20}{v-20}\right|\right)=t\)

And so, to answer the first question, we find:

\(\displaystyle t(15)=5\ln\left(\frac{21}{5}\right)\approx7.175\)

This agrees with your result, where you have taken \(v(0)=0\). Under that model, the constant of integration would indeed be zero, but you don't want to, in practice, make that assumption.

To answer the second question, I would first get velocity \(v\) as a function of time \(t\)...can you proceed?
 

hoosie

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A few steps to follow if you’re not sure how to answer the second question:8194FF9D-6D33-4D1F-BB8A-C8B3E2D86A9D.jpeg
 

MarkFL

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I would actually recommend using the exact value:

\(\displaystyle t=5\ln\left(\frac{21}{5}\right)\)

And then round \(x\) as necessary.
 

hoosie

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Yes I agree Mark - that would be better!
 
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