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A level differential equations help

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 :smile:

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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 :smile:


Can you begin solving the ODE, first of all? It's a separable equation.
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 :smile:

It's been awhile since I did this sort of stuff but I believe you have a separable differentiation equation and can change
dHdt=Hcos(0.25t)40\frac{dH}{dt}=\frac{Hcos(0.25t)}{40} to 1HdH=cos(0.25t)40dt\frac{1}{H}dH= \frac{cos(0.25t)}{40}dt by multiplying both sides of the equation by dtH\frac{dt}{H}
(edited 5 years ago)
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sqrt123
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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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sqrt123
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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
dHdt=Hcos(0.25t)40\frac{dH}{dt}=\frac{Hcos(0.25t)}{40} to 1HdH=cos(0.25t)40dt\frac{1}{H}dH= \frac{cos(0.25t)}{40}dt by multiplying both sides of the equation by dtH\dfrac{dt}{H}

Once you have \frac{1}{H}dH= \frac{cos(0.25t)}{40}dt , what do you do with that?
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 1H.dH=cos⁡(0.25t)40.dt\dfrac{1}{H}.dH = \dfrac{\cos(0.25t)}{40}.dt is for.

You can integrate both sides of this equation, yes?
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=1dyxdx=1dy then you'd say that ∫xdx=∫1dy⇒x2+C=y\int xdx= \int 1dy \Rightarrow x^2+C=y, as an example.
(edited 5 years ago)
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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 1H.dH=cos⁡(0.25t)40.dt\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=1dyxdx=1dy then you'd say that ∫xdx=∫1dy⇒x2+C=y\int xdx= \int 1dy \Rightarrow x^2+C=y, as an example.

Ahh okay, thanks :smile:

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

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.
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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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 ?
Original post by e123c
Should that be lnH = 1/160 sin(0.25t) + c ?


Nope... Differentiating RHS you get 0.25cos⁡(0.25t)160\dfrac{0.25 \cos(0.25t)}{160} which is cos⁡(0.25t)640 \dfrac{\cos(0.25t)}{640} instead.

Post your working you can't get it correct.
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Original post by RDKGames
Nope... Differentiating RHS you get 0.25cos⁡(0.25t)160\dfrac{0.25 \cos(0.25t)}{160} which is cos⁡(0.25t)640 \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 :colondollar:
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?
Original post by e123c
Oh no, it's okay I was trying to differentiate it :colondollar:
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=0.1sin⁡(0.25t)+C\ln H = 0.1 \sin(0.25t) + C

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

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

You can firstly rearrange for HH 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
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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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 :/
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 HH first since it's not that much of a stretch. We can find the constant of integration after that.
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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 HH 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) ?
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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