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I'm trying to solve this question:

MATHS

But I just can't figure out what the first step is, little help? THANKS!

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    See [Characteristic equation (calculus)](http://en.wikipedia.org/wiki/Characteristic_equation_(calculus))2011-06-19
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    ... and [Homogeneous equations with constant coefficients](http://en.wikipedia.org/wiki/Linear_differential_equation#Homogeneous_equations_with_constant_coefficients)2011-06-19
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    That $t$ is the number of _hours_, rather than _minutes_, or _years_ or whatever, was not specified in the problem. That's really sloppy.2011-06-20
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    I'm voting to close this question as off-topic because it is too specific to be of any future use.2015-01-30
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    Please type the question in instead of posting a graphic. This allows the post to load faster and helps internet searches as well as allows teachers to detect forms of cheating.2015-01-31
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    @MichaelHardy, it's possible that the text specified the units of measurement for $t$, but that the OP erased it.2015-01-31

4 Answers 4

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Your equation can be changed to $f' + 0.2 f = 0$, a homogeneous linear differential equation with constant coefficients. Try to write a polynomial whose roots will help you find a solution to your problem.

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What you have is an initial value problem: a differential equation with a particular value: $$\frac{df}{dt} = -0.2f,\qquad f(0)=100g.$$ To start, you need to solve for $f$. This is a separable equation, so just separate: $$\begin{align*} \frac{df}{dt} &= -0.2f\\ \frac{df}{f} &= -0.2\,dt. \end{align*}$$ Can you take it from here?

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If you have not yet been introduced in a formal way to differential equations, look at it this way.

You know that $$f'(t) =-0.2f(t)$$ and want to find $f(t)$.

Can you think of a function (other than the identically $0$ function) whose derivative is (always) $-0.2$ times the function?

Maybe not! But you have certainly met a function whose derivative is the function itself.

Let $g(t)=e^t$. Then $g'(t)=e^t$. More generally, if $g(t)=Ce^t$, where $C$ is a constant, then $g'(t)=g(t)$.

So $e^t$ sort of behaves like you want, except for the $-0.2$ part. Can you think of a way of modifying $e^t$ so that when you differentiate, the right constant will appear in front?

It would be best to think about this for a while by yourself, but

Hint: What happens when you differentiate $e^{kt}$, or more generally $Ce^{kt}$?

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    Ah! That actually makes a lot of sense. So -0.2 is k, right? So to find the suger remaining I'd solve for S-not in: 100=e^(-0.2*5)*S-not? Thanks for the help!2011-06-21
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    @InBetween: The **general** solution is $Ce^{-0.2t}$. At time $t=0$ there are $100$ grams, so $C=100$. To find sugar remaining after $5$ hours, substitute $t=5$ in $100e^{-0.2t}$, then use calculator. To find $t$ so that $1$ gram is left, solve $1=100e^{-0.2t}$ for $t$. To do that, take the logarithm of both sides, although to avoid negatives I would probably first rewrite as $100=e^{0.2t}$. One needs some facility with basic facts about logarithms.2011-06-21
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    @InBetween: Whatever you get as your answer to the $1$ gram question, you can check by plugging into the formula. By the way, you should get a bit over $23$ hours.2011-06-21
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If $f'(t)$ were equal to $f(t)$ what woud it be? If $f'(t)$ were equal to $-f(t)$ what woud it be? Your case is a generalization. I describe the standard approach.

Your equation can be rewritten as $y'+0.2y=0$, where $y=f(t)$. This is a homogeneous equation with constant coefficients. The standard method to solve it is by finding the roots of the caracteristic equation. In this case this equation is $X+0.2=0$. Since $X=-0.2$, the general solution is $y=Ae^{-0.2t}$. The initial condition $f(0)=100$ gives $A=100$. Thus $y=f(t)=100e^{-0.2t}$.

Let's confirm: $y=f(t)=100e^{-0.2t}$, $y'=f'(t)=100(-0.2)e^{-0.2t}=-20e^{-0.2t}$. We do have $-20e^{-0.2t}+0.2\cdot 100e^{-0.2t}\equiv 0$.