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Messages - annielam

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1
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.
$y+3y^2e^{2x}+(1+2ye^{2x})y'=0$

$M_y=1+6ye^{2x}$
$N_x=4ye^{2x}$

$R_2=\frac{M_y-N_x}{N}=\frac{1+6ye^{2x}-4ye^{2x}}{1+2ye^{2x}}=1$
$\mu=e^{\int R_2dx}=e^x$

Multiply $\mu$ to both sides
$e^{x}(y+3y^2e^{2x})+e^x(1+2ye^{2x})y'$
$M_y=e^x+e^x6ye^{2x}$
$N_x=e^x+6ye^{3x}$
Since $M_y=N_x$, $x$ is the integrating factor.

$\Phi=\int{M_x}=e^xy+y^2e^{3x}+h(y)$
$\Phi_y=e^x+2ye^{3x}+h’(y)$
$h’(y)=o$
$h(y)=C$

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

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

2
Quiz-4 / TUT0302
« on: October 18, 2019, 02:12:38 PM »
Find General Solution:
$y''+2y'+y=2e^{-t}$

$r^2+2r+r=0$
$(r+1)(r+1)=0$
$r=-1,-1$

$y=c_{1}e^{-t}+c_2te^{-t}$
$y_{p}(t)=Ae^{-t} \Rightarrow Ate^{-t} \Rightarrow At^2e^{-t}$
$=$

$y_{p}'(t)=A2te^{-t}+At^2e^{-t}$
$y_{p}''(t)=2Ae^{-t}+2Ate^{-t}+2At-e^{-t}+At^2e^{-t}$
$y_{p}''(t)=2Ae^{-t}-e^t2At-e^{-t}2At+At^2e^{-t}$
$y_{p}''(t)=2Ae^{-t}-4Ae^{-t}+At^2e^{-t}$

$Y=2Ae^{-t}-4Ate^{-t}+At^2e^{-t}+2(A2te^{-t}+At^2(-1)(e^{-t}))+At^2e^{-t}$
$Y=2Ae^{-t}-4Ate^{-t}+At^2e^{-t}+4Ate^{-t}-2At^2e^{-t}+At^2e^{-t}=2e^{-t}$

$=2Ae^{-t}=2e^{-t}$
$A=1$

$Y=C_1e^{-t}+C_2te^{-t}+t^2e^{-t}$

3
Quiz-3 / TUT0302
« on: October 11, 2019, 02:00:15 PM »
Find the Wronskian of two solutions of the given differential equation.
$x^2y''+xy'+(x^2-v^2)y=0$

$y''+\frac{x}{x^2}y'+\frac{x^2-v^2}{x^2}y=0$
$y''+\frac{1}{x}y'+{x^2+y^2}{x^2}y=0$

$p(t)=\frac{1}{x}$
$=cexp(-\int \frac{1}{x}dx)$
$=cexp(-lnx)$
$=ce^{-lnx}$
$=c\frac{1}{x}$

$W=\frac{c}{x}$

4
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.
$x^{2}y^{3}+x(1+y^{2})y'=0$

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

Multiply $\mu(\frac{1}{xy^{3}})$ to both side:
$\frac{x^{2}y^{3}}{xy^{3}}+(\frac{x+xy^{2}}{xy^{3}})y'=0$
$=x+(\frac{x}{xy^{3}}+\frac{xy^{2}}{xy^{3}})y'=0$
$=x+(\frac{1}{y^{3}}+\frac{1}{y})y'=0$
$M=x$
$N=\frac{1}{y^{3}}+\frac{1}{y}$
$My=0 = Nx=0$
Now it is exact.

Integrate M with respect to $x$ we get:
$\phi(x,y)=\frac{1}{2}x^{2}+h(y)$
Take derivative with respect to $y$ on both side:
$\phi y=N=0+h'(y)$
So $h'(y)=(\frac{1}{y^{3}}+\frac{1}{y})$
$h(y)=\int y^{-3}+y^{-1}dy$
$h(y)=-\frac{1}{4}y^{-4}+lny+C$

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

5
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 -> -∞

6
Quiz-1 / TUT0302
« on: September 27, 2019, 02:02:07 PM »
$y'-y=2te^{2t}$

$p(t)=-1$
$g(t)=2te^{2t}$
$p(t)= e^{\int -1 dt} = e^{-t}$

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

$u=2t$
$dv=e^tdt$
$du=2dt$
$v=e^t$
${\int 2te^tdt}=2te^t-{\int 2e^tdt}=2te^t-2e^t+C=2e^t(t-1)+C$

$e^{-t}y=2e^t(t-1)+C$

$y=2e^{2t}(t-1)+Ce^t$

$1=y(0)=2e^0(0-1)+Ce^0=-2+C$
$C=3$

General Solution:
$y=2e^{2t}(t-1)+3e^t$

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