Finding a function, given its derivative.

Question

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? $$\int_{0}^{50} {1\over50} dy$$ ??

Answer 1

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 $C$.

Answer 2

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

Answer 3

Like so: $\frac{dy}{y}=\frac{dt}{50}$ Now integrate both sides.