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### Topics - Yuchen Cong

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##### Quiz-4 / TUT0602 Quiz4
« on: October 19, 2019, 05:11:25 PM »
Question:
2y''+2y+1=0
Solution:
2r2+2r+1=0
r=[-2+-((-2)2-(4)(2)(1))(1/2)]/2(2)
r=[-2+-(-4)(1/2)]/4
r=-(1/2)+-(1/2)i
λ=-(1/2), µ=1/2
y(t)=C1e-(1/2)tcos(1/2)t+C2e-(1/2)tsin(1/2)t

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##### Quiz-2 / TUT0602 Quiz2
« on: October 07, 2019, 04:32:59 PM »
Q: x2y3+x(1+y2)y'=0, μ(x,y)=1/xy3

Solution:
Define that M(x,y)=x2y3, N(x,y)=x(1+y2)

Then My=3x2y2, Nx=1+y2

Since My≠Nx, so the given DE is not exact

Multiply μ(x,y)=1/xy3 to both sides:
(1/xy3)(x2y3)+(1/xy3)x(1+y2)y'=0
x+(1/y3+1/y)y'=0

Then we have M'(x,y)=x, N'(x,y)=1/y3+1/y
M'y=0, N'x=0

Since M'y=N'x, the DE is exact

Thus, there exists a function φ(x,y) such that
φx=M'(x,y) and φy=N'(x,y)

Since φx=M'(x,y)=x
Integrating both sides with respect to x, we get
φ(x,y)=(1/2)x2+h(y)

Differentiating both sides with respect to y:
φy=h'(y)

Since φy=N'(x,y)=1/y3+1/y
Then h'(y)=1/y3+1/y
h(y)=(-1/2)y-2+ln|y|+C

Thus, φ(x,y)=(1/2)x2-(1/2)y-2+ln|y|+C

Therefore, C=(1/2)x2-(1/2)y-2+ln|y|

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##### Quiz-1 / TUT0602 Quiz 1
« on: September 27, 2019, 04:05:25 PM »
Q: y'-2y=e2t, y(0)=2

Solution:

p(t)=-2, g(t)=e2t
μ(t)=e∫ -2 dt=e-2t

Multiply both sides by μ(t):
e-2ty'-2e-2ty=e2t*e-2t
(e-2ty)'=e0=1

Integrate both sides:
∫(e-2ty)'=∫ 1 dt
e-2ty=t+C, where C is a constant

Divide both sides by e-2t to isolate y:
y=(t+C)/e-2t
y=(t+C)e2t

Substituting y(0)=2:
2=(0+C)e0
2=C

Thus,
y=(t+2)e2t[/pre]

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