Lecture 21

Faculty of Science; Department of Mathematics
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Readings For This Lecture

Group 2: Add a summary of the lecture in this space. Include examples, discussion, and links to external sources, if desired.

1. Use separation of variables to solve the differential equation

$\frac{d y}{d t} = - 2 t y$

with given initial conditions $y (0) = 10$ .

Solution
seperate variables giving $1 / y d y / d t = - 2 t$ and integrate both sides by with respect to dt giving $l n (y) = - t + C$ . When $t = 0, y = 10$ thereore $C = 2.3$ . $y = e^{- t + C} = e^{- t} e^{C} = e^{2.3} e^{- t} = 10 e^{- t}$ .

2. Use separation of variables to solve the differential equation

$\frac{d y}{d t} = 1 - y$

with given initial conditions $y (0) = 0.5$ .

3. Use separation of variables to solve the differential equation

$\frac{d y}{d t} = 1 + y^{2}$

with given initial conditions $y (0) = 0.5$ .

Solution
Rearrange the equation to get: $\frac{d y}{d t} * \frac{1}{1 + y^{2}} = 1$ and integrate! $\int \frac{1}{1 + y^{2}} * \frac{d y}{d t} d t = \int 1 d t$ to get $a r c t a n (y) = t + c$ then rearrange to get: $y = t a n (t + C)$ then, substitute initial conditions $y (0) = 0.5$ to get: $C = a r c t a n (. 5)$ Plug in C to get the answer $y = t a n (t) + . 5$

4. Solve the following differential equation and initial condition, assuming that $k > 0$ is a constant:

$\frac{d y}{d t} = k y^{2 / 3}, y (0) = y_{0}$ .