Topics

1

Test 3 / MAT334 TT3 MAIN A Q2

« on: December 07, 2020, 03:46:01 PM »

Hello everyone, here is the answer of MAT334 TT3 MAIN A Q2

2

Test 3 / MAT334 TT3 MAIN A Q1

« on: December 07, 2020, 03:08:09 PM »

Hello everyone , here is the Q1 of TT3 Q1 Main A

3

Quiz-3 / tut0401 quiz 3

« on: October 11, 2019, 02:15:17 PM »

hi everyone here is the quiz question find the general solution of the given differential equation
2y'' -3y' +y = 0
therefore 2r^2 -3r + 1 = 0
therefore (2r-1)(r-1) = 0
R1 = 1/2 R2 = 1
therefore y = C1e^((1/2)t)+C2e^t

4

Quiz-2 / Tut0401 quiz 2

« on: October 04, 2019, 02:16:08 PM »

here is a quesion form quiz 2 everyone enjoy!

5

Quiz-1 / TuT0401

« on: September 27, 2019, 02:53:24 PM »

here is the solution of quiz 1

6

Quiz-1 / TUT 0401

« on: September 27, 2019, 02:00:07 PM »

Find the general solution of the given equation:
xy' = (1-y^2)^1/2
solution: Because Separable,
therefore x(dy/dx) = (1-y^2)^1/2,
Rearrange: \int(1/(1-y^2)^1/2)*dy =\int(1/x)dx, where x not equal 0, y does not equal to positive or negative 1,
therefore: arcsin(y) = ln|x| + C,
therefore general solution: y=sin(ln|x| + C) where x not equal to 0, y not equal to positive 1 or negative 1.