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### Messages - Yiran Wang

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1
##### Term Test 2 / Re: Problem 2 (morning)
« on: November 19, 2019, 01:13:26 PM »

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##### Term Test 2 / Re: Problem 2 (morning)
« on: November 19, 2019, 01:08:32 PM »

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##### Term Test 2 / Re: Problem 2 (morning)
« on: November 19, 2019, 01:08:10 PM »
This is the solution

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##### Term Test 2 / Re: Problem 4 (morning)
« on: November 19, 2019, 01:01:27 PM »

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##### Term Test 2 / Re: Problem 4 (morning)
« on: November 19, 2019, 12:58:03 PM »
This is the solution

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##### Term Test 2 / Re: Problem 3 (morning)
« on: November 19, 2019, 12:27:05 PM »
repeated root graph

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##### Term Test 2 / Re: Problem 3 (morning)
« on: November 19, 2019, 12:17:53 PM »
Det(A-rI)=0
(-2-r)(-r)+1=2r+r^2+1=0
r1=r2=-1
(A-rI)=0
we have eigenvector (1 1)
Let (A-rI)=(1 1)
We have another eignvector (-1 0)
We have x=C1e^-t *(1 1)+ C2e^-t *[ t(1 1)+(-1 0)]
e^-t u1+(te^-t – e^-t)u2=0   e^-t u1+te^-t u2=e^-t/(t^2+1)
we have u1=-t/(t^2+1)
u2=1/(t^2+1)
U1=-0.5ln(t^2+1)+C1
U2=arctant+C2
X=U1(e^-t  e^-t)+ U2(te^-t  te^-t +e^-t)

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##### Term Test 2 / Re: Problem 1 (morning)
« on: November 19, 2019, 12:01:37 PM »
r2-1=0  r2=1  r=1, -1 we have homogenous sol: y=C1e^-t+C2e^t
w=e-tet+e-tet=2  w1=-e^t  w2=e^-t
yp=e^-t *∫[-e^t * 12/(e^t+1)]/2 dt +e^-t *∫[e^-t * 12/(e^t+1)]/2 dt
=-6e^-t∫e^t /(e^t+1) dt + 6e^-t∫e^-t /(e^t+1) dt
=-6e^-t *ln(e^t+1)+6e^-t∫e^-t /(e^t+1) dt
let u=e^^t, e-t=1/u, du=e^tdt, dt=1/u du
∫e^-t /(e^t+1) dt=∫[1/u/(u+1)] *1/u du
=∫1/u2(u+1) du
=∫ (1+u-u)/ u2(u+1) du
=∫1/u2 – 1/u(u+1) du
= -1/u – lnu +ln(u+1)
=-1/e^t – lne^t+ln(^et+1)
Y= C1e^-t+C2e^t-6e^-t *ln(e^t+1) +6e^-t *(-1/e^t – t+ln(e^t+1))
Y(0)=C1+C2-6ln2-6+6ln2
Y’(0)=C1-C2-6ln2-6+6ln2
C1=0, C2=6
Y=6 e^t-6e^-t * ln(et+1) +6e^-t *(-1/e^t – t+ln(e^t+1))

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##### Quiz-5 / Lec5101 Quiz
« on: October 31, 2019, 03:07:52 PM »

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##### Quiz-4 / TUT0801 Quiz4
« on: October 18, 2019, 02:00:02 PM »

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##### Quiz-3 / TUT0801 Quiz3
« on: October 11, 2019, 02:12:15 PM »

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##### Quiz-2 / MAT244TUT0801 Quiz 2
« on: October 04, 2019, 05:45:40 PM »

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##### Quiz-1 / TUT0801 Quiz1
« on: September 27, 2019, 11:02:18 PM »

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