Lecture 21

Faculty of Science; Department of Mathematics
	Course Pages
	Course Policies
	Math Solvers
	Exams
	Quizzes
	Assignments
	Lectures

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

${\tfrac {dy}{dt}}=-2ty$

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

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

2. Use separation of variables to solve the differential equation

${\tfrac {dy}{dt}}=1-y$

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

3. Use separation of variables to solve the differential equation

${\tfrac {dy}{dt}}=1+y^{2}$

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

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

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

${\tfrac {dy}{dt}}=ky^{2/3},y(0)=y_{0}$ .