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Messages - annielam

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Term Test 1 / Re: Problem 1 (main sitting)
« on: October 23, 2019, 04:15:46 PM »
Question 1:

a) Find the integrating find and a general solution.


$\mu=e^{\int R_2dx}=e^x$

Multiply $\mu$ to both sides
Since $M_y=N_x$, $x$ is the integrating factor.


$\therefore e^xy+3y^2e^{3x}+(e^x+2ye^{3x})=C$

b) Find a solution where $y(0)=1$
Sub $y(0)=1$
$\therefore e^xy+3y^2e^{3x}+(e^x+2ye^{3x})=7$

Quiz-4 / TUT0302
« on: October 18, 2019, 02:12:38 PM »
Find General Solution:


$y_{p}(t)=Ae^{-t} \Rightarrow Ate^{-t} \Rightarrow At^2e^{-t}$
$            =$





Quiz-3 / TUT0302
« on: October 11, 2019, 02:00:15 PM »
Find the Wronskian of two solutions of the given differential equation.


$=cexp(-\int \frac{1}{x}dx)$


Quiz-2 / TUT0302
« on: October 04, 2019, 02:01:09 PM »
Problem: Show that the given equation becomes exact when multiplied by the integrating factor and solve the equation.

$My=x^{2}3y^{2} \neq Nx=\frac{1}{2}x^{2}$
Therefore it is not exact.

Multiply $\mu(\frac{1}{xy^{3}})$ to both side:
$My=0 = Nx=0$
Now it is exact.

Integrate M with respect to $x$ we get:
Take derivative with respect to $y$ on both side:
$\phi y=N=0+h'(y)$
So $h'(y)=(\frac{1}{y^{3}}+\frac{1}{y})$
$h(y)=\int y^{-3}+y^{-1}dy$

General Solution: $\frac{1}{2}x^{2}-\frac{1}{4}y^{-4}+lny=C$

Quiz-1 / Re: TUT0302
« on: September 27, 2019, 02:05:36 PM »
There is an error that I think you are missing a C value and there are three different cases for different c values.

When t -> ∞
Case 1: C = 0
By L’Hopital Rule, y -> ∞

Case 2: C > 0
Then y -> ∞

Case 3: C < 0
Then y -> -∞

Quiz-1 / TUT0302
« on: September 27, 2019, 02:02:07 PM »

$p(t)= e^{\int -1 dt} = e^{-t}$

Multiply Both Side

$e^{-t}y'-e^{-t}y = 2te^{2t}e^{-t}$
$\frac{d}{dt}$$[e^{-t}y] = 2te^{t}$
$e^{-t}y=\int 2te^tdt $

By Part

${\int 2te^tdt}=2te^t-{\int 2e^tdt}=2te^t-2e^t+C=2e^t(t-1)+C $




General Solution:

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