# A level differential equations helpWatch

#1
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
0
4 weeks ago
#2
(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.
0
4 weeks ago
#3
(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
to by multiplying both sides of the equation by
Last edited by Kvothe the Arcane; 4 weeks ago
0
#4
(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
0
#5
(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
to by multiplying both sides of the equation by
Once you have \frac{1}{H}dH= \frac{cos(0.25t)}{40}dt , what do you do with that?
0
4 weeks ago
#6
(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 is for.

You can integrate both sides of this equation, yes?
0
4 weeks ago
#7
(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 then you'd say that , as an example.
Last edited by Kvothe the Arcane; 4 weeks ago
0
#8
(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 is for.

You can integrate both sides of this equation, yes?
(Original post by Kvothe the Arcane)
Integrate both sides. If it were then you'd say that , as an example.
Ahh okay, thanks

So then I get lnH = 1/40 sin(0.25t) ?
0
4 weeks ago
#9
(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.
0
4 weeks ago
#10
(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 ...
0
#11
(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 ?
0
4 weeks ago
#12
(Original post by e123c)
Should that be lnH = 1/160 sin(0.25t) + c ?
Nope... Differentiating RHS you get which is instead.

Post your working you can't get it correct.
0
#13
(Original post by RDKGames)
Nope... Differentiating RHS you get which is 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?
0
4 weeks ago
#14
(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

You can firstly rearrange for before applying the initial condition. Can you do that?
0
#15
(Original post by RDKGames)
Don't forget the +C.

So we have

You can firstly rearrange for 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
0
4 weeks ago
#16
(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 ??
0
#17
(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 :/
0
4 weeks ago
#18
(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 first since it's not that much of a stretch. We can find the constant of integration after that.
0
#19
(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 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) ?
0
4 weeks ago
#20
(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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