A level differential equations help Watch

e123c
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Hey,

I'm stuck on a differential equations question and don't really know how to get started:

The height above ground, H metres, of a child on a fairground ride can be modelled by the differential equation

dH/dt = (Hcos(0.25t)) / 40

where t is the time, in seconds, from the start of the ride.

Given that the child is 5 m above ground at the start of the ride, show that H = 5 e^(0.1sin(0.25t)).

I'd really appreciate any help
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RDKGames
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(Original post by e123c)
Hey,

I'm stuck on a differential equations question and don't really know how to get started:

The height above ground, H metres, of a child on a fairground ride can be modelled by the differential equation

dH/dt = (Hcos(0.25t)) / 40

where t is the time, in seconds, from the start of the ride.

Given that the child is 5 m above ground at the start of the ride, show that H = 5 e^(0.1sin(0.25t)).

I'd really appreciate any help
Can you begin solving the ODE, first of all? It's a separable equation.
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Kvothe the Arcane
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(Original post by e123c)
Hey,

I'm stuck on a differential equations question and don't really know how to get started:

The height above ground, H metres, of a child on a fairground ride can be modelled by the differential equation

dH/dt = (Hcos(0.25t)) / 40

where t is the time, in seconds, from the start of the ride.

Given that the child is 5 m above ground at the start of the ride, show that H = 5 e^(0.1sin(0.25t)).

I'd really appreciate any help
It's been awhile since I did this sort of stuff but I believe you have a separable differentiation equation and can change
\frac{dH}{dt}=\frac{Hcos(0.25t)}  {40} to \frac{1}{H}dH= \frac{cos(0.25t)}{40}dt by multiplying both sides of the equation by \frac{dt}{H}
Last edited by Kvothe the Arcane; 4 weeks ago
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e123c
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(Original post by RDKGames)
Can you begin solving the ODE, first of all? It's a separable equation.
As in, integrate dH/dt ? Because I'm not sure how I should do that
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e123c
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(Original post by Kvothe the Arcane)
It's been awhile since I did this sort of stuff but I believe you have a separable differentiation equation and can change
\frac{dH}{dt}=\frac{Hcos(0.25t)}  {40} to \frac{1}{H}dH= \frac{cos(0.25t)}{40}dt by multiplying both sides of the equation by \dfrac{dt}{H}
Once you have \frac{1}{H}dH= \frac{cos(0.25t)}{40}dt , what do you do with that?
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RDKGames
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(Original post by e123c)
As in, integrate dH/dt ? Because I'm not sure how I should do that
Not quite. You want to integrate an expression in terms of H multiplied by dH, and an expression in terms of t multiplied by dt.

This is precisely what the form \dfrac{1}{H}.dH = \dfrac{\cos(0.25t)}{40}.dt is for.

You can integrate both sides of this equation, yes?
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Kvothe the Arcane
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(Original post by e123c)
Once you have \frac{1}{H}dH= \frac{cos(0.25t)}{40}dt , what do you do with that?
Integrate both sides. If it were xdx=1dy then you'd say that \int xdx= \int 1dy \Rightarrow x^2+C=y, as an example.
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e123c
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(Original post by RDKGames)
Not quite. You want to integrate an expression in terms of H multiplied by dH, and an expression in terms of t multiplied by dt.

This is precisely what the form \dfrac{1}{H}.dH = \dfrac{\cos(0.25t)}{40}.dt is for.

You can integrate both sides of this equation, yes?
(Original post by Kvothe the Arcane)
Integrate both sides. If it were xdx=1dy then you'd say that \int xdx= \int 1dy \Rightarrow x^2+C=y, as an example.
Ahh okay, thanks

So then I get lnH = 1/40 sin(0.25t) ?
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RDKGames
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(Original post by e123c)
Ahh okay, thanks

So then I get lnH = 1/40 sin(0.25t) ?
Check your RHS. It doesn't *quite* differentiate to cos(0.25t)/40 therefore you are off by a constant.

Also, you are forgetting to include the constant of integration.
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Muttley79
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(Original post by e123c)
Once you have \frac{1}{H}dH= \frac{cos(0.25t)}{40}dt , what do you do with that?
Integrate both sides ...
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e123c
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(Original post by RDKGames)
Check your RHS. It doesn't *quite* differentiate to cos(0.25t)/40 therefore you are off by a constant.

Also, you are forgetting to include the constant of integration.
Should that be lnH = 1/160 sin(0.25t) + c ?
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RDKGames
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(Original post by e123c)
Should that be lnH = 1/160 sin(0.25t) + c ?
Nope... Differentiating RHS you get \dfrac{0.25 \cos(0.25t)}{160} which is  \dfrac{\cos(0.25t)}{640} instead.

Post your working you can't get it correct.
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e123c
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(Original post by RDKGames)
Nope... Differentiating RHS you get \dfrac{0.25 \cos(0.25t)}{160} which is  \dfrac{\cos(0.25t)}{640} instead.

Post your working you can't get it correct.
Oh no, it's okay I was trying to differentiate it
Now I get lnH = 0.1sin(0.25t) which makes a lot more sense, do I then just show H = e^(0.1sin(0.25t)) ? And then just put the 5 in as that's the initial value or what should I do about that bit?
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RDKGames
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(Original post by e123c)
Oh no, it's okay I was trying to differentiate it
Now I get lnH = 0.1sin(0.25t) which makes a lot more sense, do I then just show H = e^(0.1sin(0.25t)) ? And then just put the 5 in as that's the initial value or what should I do about that bit?
Don't forget the +C.

So we have \ln H = 0.1 \sin(0.25t) + C

You can firstly rearrange for H before applying the initial condition. Can you do that?
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e123c
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(Original post by RDKGames)
Don't forget the +C.

So we have \ln H = 0.1 \sin(0.25t) + C

You can firstly rearrange for H before applying the initial condition. Can you do that?
Do you just make the +c equal to 5? I'm really lost on how to incorporate the 5
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RDKGames
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(Original post by e123c)
Do you just make the +c equal to 5? I'm really lost on how to incorporate the 5
How did you get c = 5 ??
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e123c
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(Original post by RDKGames)
How did you get c = 5 ??
Just on the basis that the initial height is 5 and I know I need to add it somewhere :/
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RDKGames
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(Original post by e123c)
Just on the basis that the initial height is 5 and I know I need to add it somewhere :/
That's not a very good reason.

Come on, reason it mathematically. But what I'm suggesting if that you first ignore the initial conditions. You should rearrange for H first since it's not that much of a stretch. We can find the constant of integration after that.
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e123c
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(Original post by RDKGames)
That's not a very good reason.

Come on, reason it mathematically. But what I'm suggesting if that you first ignore the initial conditions. You should rearrange for H first since it's not that much of a stretch. We can find the constant of integration after that.
Is that just H = e^(0.1sin(0.25t) + c) ?
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Kvothe the Arcane
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(Original post by e123c)
Do you just make the +c equal to 5? I'm really lost on how to incorporate the 5
Initial height is 5. I.e. H is 5, when t=0 but you need to have a workable equation to figure out what the constant of integration is first.
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