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At the start of day zero during the summer, the temperature is $y(0) = 15$ degrees Celsius. Over a 50 day period the temperature increases according to the rule $$y'(t) = {y(t) \over 50}$$. With time $t$ measured in days. Find the formula for $y$

I not sure how to start here. Can I integrate $y'(t)$ to get back $y(t)$ ? Would it have to be a definite integral from zero to fifty ? Or and indefinite? $$\int_{0}^{50} {1\over50} dy$$ ??

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  • $\begingroup$ Separate the variables then integrate $\endgroup$ – randomgirl Apr 21 '16 at 13:00
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Let's rewrite $y'(t)$ as $\dfrac{dy}{dt}$ and just write $y(t)$ as $y$. Then we have: $$ \frac{dy}{dt} = \frac{1}{50} y$$

Separate variables to get: $$ \frac{dy}{y} = \frac{1}{50} \, dt$$

Integrate both sides: \begin{align} \int \frac{dy}{y} &= \int \frac{1}{50} \, dt\\[0.3cm] \ln |y| &= \frac{1}{50} t + C\\[0.3cm] |y| &= e^{(t/50) + C}\\[0.3cm] y &= Ce^{t/50} \end{align}

Now use the fact that $y(0) = 15$ to find $C$.

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Hint. Note that: $$y'-\frac{1}{50}y=0\\ e^{-t/50}y'-\frac{1}{50}e^{-t/50}y=0\\ \big(e^{-t/50}y(t)\big)'=0\\ e^{-t/50}y(t)=\text{constant}$$

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    $\begingroup$ The product rule for differentiation is probably the best way to solve this for someone who's not used to the tools of differential equations (and since the asker is stumped by this question, I would assume that that is the case). Although going from the second to third line by multiplying with $e^{-t/50}$ might seem like some sort of magic trick pulled up from a hat, that's what makes it work in the end. $\endgroup$ – Arthur Apr 21 '16 at 13:03
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Like so: $\frac{dy}{y}=\frac{dt}{50}$ Now integrate both sides.

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