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Help with Core 4 Differential problem

Hi,

I thought I had this figured out, but I am just doing past papers and the question is

The height, h metres, of a shrub t years after planting is given by the differential equation

dh/dt = 6-h/20

A shrub is planted when its height is 1m.

(i) Show by integration that t = 20ln(5/6-h)

That question I'm having trouble with. I know you have to turn each side over to get

dt/dh = 20/6-h

And then integrate that to get t=, but I thought the integral of 1/ax+b was

1/a ln(ax + b)

which in this case would give 20ln(6-h) I thought?

So I'm not too sure where they got the 5/ bit from. If someone could show me how they reach that point I'd be grateful thanks. :smile:

Reply 1

TheTallOne

You are forgetting to add your constant.

Remember when first planted, the shrub is 1m tall.

Reply 2

Swayum

You're also forgetting to divide by -1 since it's -1h + 6.

Reply 3

moves-my-world

dh/6-h = dt/20
ln |6-h| = 1/20t + c
6-h = e^(1/20t + c)
6-h = e^1/20t X e^c
let e^c = A (or whatever letter you choose)
6-h = Ae^1/20t
6-Ae^1/20t = h

This help?

Reply 4

Lou Reed

you need to seperate the variables too.

Reply 5

Lou Reed

moves-my-world

dh/6-h = dt/20
ln |6-h| = 1/20t + c
6-h = e^(1/20t + c)
6-h = e^1/20t X e^c
let e^c = A (or whatever letter you choose)
6-h = Ae^1/20t
6-Ae^1/20t = h

This help?

you cant do the bolded line, you need to turn the c into a lnC first and combine the two lns on the RHS before you put it all to the power of e.

Reply 6

Swayum

Lou Reed

you cant do the bolded line, you need to turn the c into a lnC first and combine the two lns on the RHS before you put it all to the power of e.

You can do it his way. Writing c as lnC is just a more explicit way of doing it rather than doing e^(x + c) = Ae^x.

Reply 7

Lou Reed

what i was saying is that you cant do it because there is a "+", thats what i think anyways..

Reply 8

Swayum

He's exponentiated both sides of the equation. After that, he's split it using C1 knowledge that a^(m + n) = a^m * a^n.

Reply 9

username165864

Lou Reed

what i was saying is that you cant do it because there is a "+", thats what i think anyways..

x^2 \times x^3 = x^5

EDIT: Swayum... You are too quick! :smile:

Reply 10

Lou Reed

ohh..hmm..so i can do like
ln|1+x| = y/6 + 4 + 5y
as 1+x = e^(y/6 + 4 + 5y) ?

btw for the question asked i dont think you need that anyway =/

Reply 11

gee0396

when you state dh/6-h, do you mean (1/6-h) .dh or not? i'm currently doing the original question and am really confused

Reply 12

gee0396

Original post by gee0396

when you state dh/6-h, do you mean (1/6-h) .dh or not? i'm currently doing the original question and am really confused

Reply 13

gee0396

would you mind putting up your full answer to this question please? am also currently stuck and would love to see what you do!!! :s-smilie:

Reply 14

davros

Original post by gee0396

would you mind putting up your full answer to this question please? am also currently stuck and would love to see what you do!!! :s-smilie:

This question is nearly 5 years old now, so the original posters have probably gone away!

Anyway,

the original equation is

\displaystyle \dfrac{dh}{dt} = \dfrac{6-h}{20}

.

If you separate this and integrate you get to:

\displaystyle \int \dfrac{dh}{6-h} = \int \dfrac{dt}{20}

Can you continue from here?

Reply 15

Swayum

Original post by davros

This question is nearly 5 years old now, so the original posters have probably gone away!

Pretty embarrassing for me...

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