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applying differential equations to real-life problems watch

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    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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    (Original post by Alen.m)
    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
    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: \displaystye \int -k \, \mathrm{d}t = -k \int\, \mathrm{d}t = -kt + c
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    (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: \displaystye \int -k \, \mathrm{d}t = -k \int\, \mathrm{d}t = -kt + c
    Thanks man
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    (Original post by Alen.m)
    Thanks man
    No problem.
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    (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
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    (Original post by samb1234)
    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
    You are. No problem about it.
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    (Original post by Zacken)
    You are. No problem about it.
    thanks, i hate using +c lol
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    (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 AName:  image.jpeg
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    (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 AName:  image.jpeg
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    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 or 2x-c or 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.
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    (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 or 2x-c or 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 Name:  image.jpg
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    (Original post by Alen.m)
    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 Name:  image.jpg
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    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.
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    (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 Name:  image.jpg
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    (Original post by Alen.m)
    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 Name:  image.jpg
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    I think the textbook is wrong, tbh. But really, all you need to know is that 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 as t increases.
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    (Original post by Zacken)
    I think the textbook is wrong, tbh. But really, all you need to know is that 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 as 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
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    (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.
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    (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?
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    (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.
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    (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?
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    (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?
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    (Original post by Zacken)
    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?
    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 Attachment 511585511587
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