MAT244--2019F > Term Test 1

Problem 3 (morning)

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GuangyuDu:
Question 3:
$
y''-6y'+8y=48\sin h (2x), y(0)=0, y'(0)=0.
$

Solution:
$
r^2-6r+8=0
$
$r=4, r=2$
$y_c(x)=c_1e^{4t}+c_2e^{2t}$
$48\sin h (2x)=24\cdot e^{2x}-24\cdot e^{-2x}$
$
y_p(x)=Axe^{2x}, y_p'=2Axe^{2x}+Ae^{2x},
$
$
y_p''(x)=4Axe^{2x}+2Ae^{2x}
$
$
y''-6y'+8y=-2Ae^{2x}=24e^{2x}
$
$
A=-12,y_p(x)=-12xe^{2x}
$
$
y_p(x)=Be^{-2x},y_p'(x)=-2Be^{-2x},y_p''(x)=4Be^{-2x}
$
$
y''-6y'+8y=24Be^{-2x}=-24e^{-2x}
$
$
B=-1,y_p(x)=-e^{-2x}
$
$
y=y_c+y_p=c_1e^{4t}+c_2e^{2t}-12xe^{2x}-e^{-2x}
$

Plug in $y(0)=0$, $y'(0)=0$.

we have $c_1=4,c_2=3$
$
y=4e^{4t}+3e^{2t}-12xe^{2x}-e^{-2x}
$

Carrie:
$$y''-6y'+8y=48sinh(2x)$$
$$y(0)=0 \ y'(0)=0 \ sinhx=\frac{{e^x-e^{-x}}}{2}$$
$r^2-6r+8=0$\\
(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\\
$y_2(x)=-e^{-2x}$\\
$y(x)=c_1e^{2x}+c_2e^{4x}-12xe^{2x}-e^{-2x}$\\
$y(0)=c_1+c_2-1=0$\\
$y'(x)=2c_1e^{2x}+4c_2e^{4x}-24xe^{2x}-12e^{2x}+2e^{-2x}$\\
$y'(0)=2c_1+4c_2-12+2=0$\\
$c_1=-3 \ c_2=4$\\
$y(x)=-3e^{2x}+4e^{4x}-12xe^{2x}-e^{-2x}$

Xuefeng Fan:
Hi everyone here is the solution of question 3:
y''−6y'+8y=48sinh(2x),y(0)=0,y'(0)=0.

Solution:
R^2−6R+8=0
 therefore R=4,R=2
y=C1e^4t+c2e^2t

yp=Axe^2x,y'=2Ae^2x+Ae^2𝑥,
y''=4Axe^2𝑥+2Ae^2x
A=−12,Y=−12xe^2x
y=Be^−2x,
y'=−2Be^−2x,
y''=4Be^−2x
plug in
B=−1,yp=−e^−2x
Y=C1e^4t+C2e^2t−12Xe^x−e−2x

Plug in the value

we have C1=4,C2=-3

Xinyu Jing:
𝑦′′−6𝑦′+8𝑦=48sinh(2𝑥)

𝑦′′−6𝑦′+8𝑦=$24𝑒^{2𝑥}−24𝑒^{−2𝑥}$

 Let $𝑦=𝑒^{𝑟𝑥(2)}$

𝑦′=sec(𝑟𝑥) 𝑦′′=$𝑟^{2}𝑒^{𝑟𝑥(3)}$

$r^{2}−r𝑥+8=0$

$𝑟_{1}=2, 𝑟_{2}=4$

$𝑦=𝑐_{1}𝑒^{4𝑥}+𝑐_{2}𝑒^{2x}$

 let 𝑦=$𝐴𝑥𝑒^{2𝑥}$

𝑦′=$𝐴𝑒^{2𝑥}+2𝐴𝑥𝑒^{2𝑥}$

𝑦″=$4𝐴𝑒^{2𝑥}𝑥+2𝐴𝑒^{2𝑥}+2𝐴𝑒^{2𝑥}=4𝐴𝑒^{2𝑥}𝑥+4𝐴𝑒^{2𝑥}$

$(8𝐴−12𝐴+4𝐴)𝑡⋅𝑒^{2𝑥}+(−6𝐴+4𝐴)𝑒^{2𝑥}=24𝑒^{2𝑥}−2𝐴𝑒^{2𝑥}=24𝑒^{2𝑥}$

𝐴=−12 $𝑦=−12𝑥𝑒^{2𝑥}$

 Let 𝑦=$B𝑒^{−2𝑥}$

𝑦′=$−2B𝑒^{−2𝑥}$

𝑦′′=$4B𝑒^{−2𝑥}$

$4𝐵𝑒^{−2𝑥}+12𝐵𝑒^{−2𝑥}+8𝐵𝑒^{−2𝑥}=−24𝑒^{−2𝑥}$

$24𝐵𝑒^{−2𝑥}=−24𝑒^{−2𝑥}$

𝐵=−1

∴$𝑦𝑝(𝑥)=−𝑒^{−2𝑥}$

so
$𝑦=𝑎𝑒^{2}+𝑐_{2}𝑒^{4𝑥}−12𝑒^{2𝑥}−𝑒^{−2𝑥}$

$𝑦′=2𝑐_{1}𝑒^{2𝑡}+4𝑐_{2}𝑒^{4𝑡}−24𝑟𝑒^{2𝑥}+2𝑒^{−2𝑥}−12𝑒^{2𝑥}$

𝑦(0)=𝑦′(0)=0

$𝑦=−3𝑒^{2𝑥}+4𝑒^{4𝑥}−12𝑥𝑒^{2𝑥}−𝑒^{−2𝑥}$

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