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Messages - Carrie

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1
Term Test 1 / Re: Problem 3 (morning)
« on: October 23, 2019, 11:12:34 AM »
$$y''-6y'+8y=48sinh(2x)$$
$$y(0)=0 \ y'(0)=0 \ sinhx=\frac{{e^x-e^{-x}}}{2}$$
$r^2-6r+8=0$\\
(r-2)(r-4)=0\\
r=2 \ r=4 \\
$y(x)=c_1e^{2x}+c_2e^{4x}$\\
$48sinh(2x)=48(\frac{e^x-e^{-x}}{2})=48(\frac{e^{2x}}{2}-\frac{e^{-2x}}{2})=24e^{2x}-24e^{-2x}$\\
$y"-6y'+8y=24e^{2x}$\\
$y_1(x)=Axe^{2x}$\\
$y_1'(x)=2Axe^{2x}+Ae^{2x}$\\
$y_{1}''(x)=4Axe^{2x}+4Ae^{2x}$\\
Substitute \  them\  back \ to \ the \ original \ equation\\
$4Axe^{2x}+4Ae^{2x}-12Axe^{2x}-6Ae^{2x}+8Axe^{2x}=24e^{2x}$\\
$-2Ae^{2x}=24Ae^{2x}$\\
A=-12\\
$y_1(x)=-12xe^{2x}$\\
$y''-6y'+8y=-24e^{-2x}$\\
$y_2(x)=Ae^{-2x}$\\
$y_2(x)=-2Ae^{-2x}$\\
$y_2(x)=4Ae^{-2x}$\\
Substitute \  them\  back \ to \ the \ original \ equation\\
$4Ae^{-2x}+12-2Ae^{-2x}+8Ae^{-2x}=-24Ae^{-2x}$\\
$24Ae^{-2x}=-24e^{-2x}$\\
A=-1\\
$y_2(x)=-e^{-2x}$\\
$y(x)=c_1e^{2x}+c_2e^{4x}-12xe^{2x}-e^{-2x}$\\
$y(0)=c_1+c_2-1=0$\\
$y'(x)=2c_1e^{2x}+4c_2e^{4x}-24xe^{2x}-12e^{2x}+2e^{-2x}$\\
$y'(0)=2c_1+4c_2-12+2=0$\\
$c_1=-3 \ c_2=4$\\
$y(x)=-3e^{2x}+4e^{4x}-12xe^{2x}-e^{-2x}$

2
Quiz-3 / TUT 0501
« on: October 11, 2019, 04:19:34 PM »
y’’+4y’+3y=0     y(0)=2    y’(0)=-1
r^2+4r+3=0
(r+3)(r+1)=0
r=-3   r=-1
y(t)=c1e^-3t+c2e^-t
Since y(0)=2
y(0)=c1+c2=2
c2=2-c1                        (1)
y’(t)=-3c1e^-3t-c2e^-t
y’(0)=-1
y’(0)=-3c1-c2=-1
-3c1-(2-c1)=-1
-3c1-2+c1=-1
-2c1=1
c1=-1/2
c2=5\2                from(1)
y(t)=-1\2e^-3t+5\2e^-t

3
Quiz-1 / TUT0501
« on: September 27, 2019, 04:29:57 PM »
Solution

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