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##### Quiz-4 / TUT 0501
« on: October 18, 2019, 02:15:33 PM »
y”+2y’+2y=0.  y(π/4)=2.      y’(π/4)=-2。        （note： π is “pie”）

r^2+2r+2=0
r1=-1+i
r2=-1-i

y=C1e^(-t)cos(t)+C2e^(-t)sin(t)

y(π/4)=2=[e^(-π/4)](1/√2)(C1+C2)
C1+C2=2/([e^(-π/4)] (1/√2))

y’(t)=-C1[e^(-t)]cos(t)-C1[e^(-t)]sin(t)
-2=-2[e^(-π/4)](1/√2)C1
C1=√2/[e^(π/4)]=C2
y=(√2/[e^(π/4)])*e^(-t)cos(t)+(√2/[e^(π/4)])*e^(-t)sin(t)

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##### Quiz-3 / Tut 0501
« on: October 11, 2019, 01:53:59 PM »
(t^2)y”-t(t+2)y’+(t+2)y=0
y”+(-1)(t)(t+2)(y’)/(t^2)+(t+2)(y)/(t^2)
P(t)=-(t+2)/t=-(1+(2/t))
W=ce^(-∫P(t)dt) =ce^(∫1+(2/t)dt) =c(e^t)*(e^(ln(t^2)) =c(t^2)(e^t)

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##### Quiz-1 / TUT0501
« on: September 27, 2019, 10:07:39 AM »
Q:   ty’ +(t+1)y=t. y(ln(2))=1
Ans:
y’+[(t+1)/t]y=1
p(t)=[(t+1)/t]
u=e^(∫[(t+1)/t]dt)=te^t
te^t*y’+[(t+1)/t]*te^t*y=te^t
te^t *y=∫te^tdt
u=t   du=1dt
v=e^t   dv=e^tdt
te^t *y = te^t-∫e^t dt = te^t - e^t +c
y=1-(1/t)+(c/(te^t))

y=1 t=ln2
1=1-1/ln2 +c/(ln2)
C=2
y=1-1/t+2/(te^t)

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