applying differential equations to real-life problems

Hi guys ,
I'm practising differential equations where you need to integrate both sides to find the general solution as it's shown in the attachment I'm a bit confused of integrating -k dt which which gives out -kt can anyone help?thanks

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Reply 1

Zacken

Original post by Alen.m

k

is a constant, i.e: nothing but a number. How would you integrate

\int -1 \, \mathrm{d}t

? You'd get

-t +c

, I hope?

So:

Unparseable latex formula:

\displaystye \int -k \, \mathrm{d}t = -k \int\, \mathrm{d}t = -kt + c

(edited 8 years ago)

Reply 2

Alen.m

Original post by Zacken

k

is a constant, i.e: nothing but a number. How would you integrate

\int -1 \, \mathrm{d}t

? You'd get

-t +c

, I hope?

So:

Unparseable latex formula:

\displaystye \int -k \, \mathrm{d}t = -k \int\, \mathrm{d}t = -kt + c

Thanks man

Reply 3

Zacken

Original post by Alen.m

Thanks man

No problem.

Reply 4

samb1234

Original post by Zacken

No problem.

On the topic of DE's, do you know if i'm allowed to integrate between limits rather than using +c in c4 (e.g. integrate between p0 and P), which is mathematically valid but i don't think it is normally in mark schemes

Reply 5

Zacken

Original post by samb1234

You are. No problem about it.

Reply 6

samb1234

Original post by Zacken

You are. No problem about it.

thanks, i hate using +c lol

Reply 7

Alen.m

Original post by Zacken

You are. No problem about it.

Can i just also ask you one single question about e equations in the attachment, im abit confused and also couldnt find anything usefull on my note about how 1/a is written as A

Reply 8

Zacken

Original post by Alen.m

Can i just also ask you one single question about e equations in the attachment, im abit confused and also couldnt find anything usefull on my note about how 1/a is written as A

A

is an arbitrary constant. That is, it can be any number. It doesn't matter what form I like it in.

For example, if I was doing

\int 2 \, \mathrm{d}x = 2x + c

, I could write the

c

in whichever way I want. I could use

2x + c^2

2x-c

2x + A

, it doesn't matter and nobody cares because it'll all work out in the end.

So, in cases like this. you could leave your answer as

\frac{1}{a}e^{-kt}

but we choose to say

A = \frac{1}{a}

and write

Ae^{-kt}

because it just looks nicer.

Reply 9

Alen.m

Original post by Zacken

A

is an arbitrary constant. That is, it can be any number. It doesn't matter what form I like it in.

For example, if I was doing

\int 2 \, \mathrm{d}x = 2x + c

, I could write the

c

in whichever way I want. I could use

2x + c^2

2x-c

2x + A

, it doesn't matter and nobody cares because it'll all work out in the end.

So, in cases like this. you could leave your answer as

\frac{1}{a}e^{-kt}

but we choose to say

A = \frac{1}{a}

and write

Ae^{-kt}

because it just looks nicer.

All clear now, for the other part i was trying to sketch the graph of y=30000-20000e^0.2In0.9t and saw this bullet point from the text book saying that remmeber that0.2In0.9t is negative , can you please touch on that a little bit so i underestand what the text book meant by that thanks

Reply 10

Zacken

Original post by Alen.m

Are you meant to sketch the graph in some interval? Because that's quite wrong.

0.2 \ln (0.9 \times 2) \approx 0.1 > 0

for example.

Reply 11

Alen.m

Original post by Zacken

Are you meant to sketch the graph in some interval? Because that's quite wrong.

0.2 \ln (0.9 \times 2) \approx 0.1 > 0

for example.

Well actually it's part c of the question in the attachment , i've managed to do the rest of it but kind of confused by the given hint by text book on part c. I can send you all answers as attachemnets if you want so you dont go through all of them

Reply 12

Zacken

Original post by Alen.m

I think the textbook is wrong, tbh. But really, all you need to know is that

Unparseable latex formula:

e^{\blah}

is always positive, so you're always doing

30 000 - \text{something positive that gets smaller and smaller}

so the graph nears the line

y =30,000

t

increases.

Reply 13

Alen.m

Original post by Zacken

I think the textbook is wrong, tbh. But really, all you need to know is that

Unparseable latex formula:

e^{\blah}

is always positive, so you're always doing

30 000 - \text{something positive that gets smaller and smaller}

so the graph nears the line

y =30,000

t

increases.

It's been couple of times that its answer makes me confus and lost tbh but as you said all i need to do is to find the 10000 value on c graph first and the line nears 30000 as t increases

(edited 8 years ago)

Reply 14

Zacken

Original post by Alen.m

It's been couple of times that its answer makes me confus and lost tbh but as you said all i need to is to find the 10000 value on c graph first and the line nears 30000 as t increases

Yep, pretty much.

Reply 15

Alen.m

Original post by Zacken

Yep, pretty much.

Actually the graph suggests that c gets bigger and bigger as it approaches 30000 and as t increases isn't?

Reply 16

Zacken

Original post by Alen.m

Actually the graph suggests that c gets bigger and bigger as it approaches 30000 and as t increases isn't?

Ah, fuark! Ignore me. I thought the equation was

\ln (0.9 t)

instead of

t\ln (0.9)

, So what we have is

e^{-0.02t}

approximately, so as t gets bigger,

e^{-0.02t}

gets smaller and smaller, so the graph approaches

30 000

. My apologies.

Reply 17

Alen.m

Original post by Zacken

Ah, fuark! Ignore me. I thought the equation was

\ln (0.9 t)

instead of

t\ln (0.9)

, So what we have is

e^{-0.02t}

approximately, so as t gets bigger,

e^{-0.02t}

gets smaller and smaller, so the graph approaches

30 000

. My apologies.

But c on the graph is increasing as the graph suggests isn't? Is it because it's been muliplied by 20000?

Reply 18

Zacken

Original post by Alen.m

But c on the graph is increasing as the graph suggests isn't? Is it because it's been muliplied by 20000?

It's because you're doing 30,000 - something that gets smaller and smaller. So as you start out, it's quite big and 30,000 - big number = small number. But then as t gets bigger, e^(-0.02t) gets smaller so you do 30,000 - smaller number = bigger number. And hence the graph is increasing. But then, at a certain point, when t gets really big e^(-0.02t) is almost 0 so 30,000 - almost 0 is almost 30,000, etc...? You get me?

Reply 19

Alen.m

Original post by Zacken

So basically it's a graph of y=30000-20000e^0.2tIn(0.9)? The reason im asking is because the text book stated it's a graph of e^0.2In(0.9t) which should definetely be different

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