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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
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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
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Test 2 / Re: possible typo on 2020 Question 3 Morning sitting
« on: December 07, 2020, 02:46:20 PM »
Yes I am totally agree with you
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Test 2 / Re: Question 3c for 2020S Night Sitting
« on: December 07, 2020, 02:43:43 PM »
f(x,y) = coshxsiny +i(-sinhxcosy+c)
=cos(ix)siny +i(-(-isin(ix)cosy +c)
=cos(ix)siny - sin(ix)cosy +ic
=sin(y-ix)+ic
=sin(-i(x+iy))+ic
=-sin(i(x+iy)) + ic
=-isinh(x+iy) +ic
=-isinhz+ic
=cos(ix)siny +i(-(-isin(ix)cosy +c)
=cos(ix)siny - sin(ix)cosy +ic
=sin(y-ix)+ic
=sin(-i(x+iy))+ic
=-sin(i(x+iy)) + ic
=-isinh(x+iy) +ic
=-isinhz+ic
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Test 2 / Re: Possible typo in 2018 Version B Q3(b)
« on: December 07, 2020, 02:34:56 PM »
After find what is Vx and Vy, Vy=Ux Vx=-Uy
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Test 1 / Re: 2020 Night Sitting #1
« on: December 07, 2020, 02:28:14 PM »
After finding out that e^z=-\frac{1}{2}\:i\frac{\sqrt{3}}{2}
We can take ln and get the answer that z =i(+- (2/3)pi +2kpi)
We can take ln and get the answer that z =i(+- (2/3)pi +2kpi)
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Term Test 1 / Re: Problem 3 (morning)
« on: October 23, 2019, 03:50:41 PM »
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
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
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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
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
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Quiz-2 / Tut0401 quiz 2
« on: October 04, 2019, 02:16:08 PM »
here is a quesion form quiz 2 everyone enjoy!
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